NCERT Class 10 Maths Solutions: Quadratic Equations
Question:
Find two consecutive positive integers, sum of whose squares is 365.
Concept in a Minute:
Quadratic Equations: Understanding how to form and solve quadratic equations from word problems. This involves setting up an equation with a variable representing the unknown integer(s) and then manipulating it into the standard quadratic form ax^2 + bx + c = 0. Factoring or using the quadratic formula are key methods for solving these equations.
Explanation:
Let the first positive integer be x.
Since the integers are consecutive, the next positive integer will be x + 1.
The problem states that the sum of their squares is 365.
So, we can write the equation:
x^2 + (x + 1)^2 = 365
Expand the term (x + 1)^2:
(x + 1)^2 = x^2 + 2*x*1 + 1^2 = x^2 + 2x + 1
Substitute this back into the equation:
x^2 + (x^2 + 2x + 1) = 365
Combine like terms:
2x^2 + 2x + 1 = 365
Subtract 365 from both sides to set the equation to zero (standard quadratic form):
2x^2 + 2x + 1 – 365 = 0
2x^2 + 2x – 364 = 0
We can simplify this quadratic equation by dividing all terms by 2:
x^2 + x – 182 = 0
Now, we need to solve this quadratic equation for x. We can try to factor it. We are looking for two numbers that multiply to -182 and add up to +1.
Let’s list factors of 182:
1 * 182
2 * 91
7 * 26
13 * 14
The pair 13 and 14 are close. If we use +14 and -13, their product is -182 and their sum is +1.
So, we can factor the equation as:
(x + 14)(x – 13) = 0
This gives us two possible solutions for x:
x + 14 = 0 => x = -14
x – 13 = 0 => x = 13
The question asks for two consecutive *positive* integers. Therefore, we discard the negative solution x = -14.
The first positive integer is x = 13.
The second consecutive positive integer is x + 1 = 13 + 1 = 14.
Let’s check our answer:
Sum of squares = 13^2 + 14^2
= 169 + 196
= 365
The sum of the squares is indeed 365.
The two consecutive positive integers are 13 and 14.
Question:
Find the roots of the following quadratic equation by factorisation:
$\sqrt{2𝑥2+7𝑥+5\sqrt{2=0}$
Concept in a Minute:
The core concept is factorisation of quadratic equations. This involves finding two numbers that multiply to the product of the first and last term coefficients and add up to the middle term coefficient. These two numbers are then used to split the middle term, allowing for grouping and factoring.
Explanation:
The given quadratic equation is $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.
Step 1: Identify the coefficients.
Here, $a = \sqrt{2}$, $b = 7$, and $c = 5\sqrt{2}$.
Step 2: Find the product of $a$ and $c$.
$a \times c = \sqrt{2} \times 5\sqrt{2} = 5 \times (\sqrt{2} \times \sqrt{2}) = 5 \times 2 = 10$.
Step 3: Find two numbers that multiply to 10 and add up to $b$ (which is 7).
The two numbers are 5 and 2, because $5 \times 2 = 10$ and $5 + 2 = 7$.
Step 4: Split the middle term ($7x$) using these two numbers.
$\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$.
Step 5: Group the terms and factor.
Group the first two terms and the last two terms:
$(\sqrt{2}x^2 + 5x) + (2x + 5\sqrt{2}) = 0$.
Factor out the common factor from each group.
From the first group, $x$ is common: $x(\sqrt{2}x + 5)$.
From the second group, $\sqrt{2}$ is common: $\sqrt{2}( \sqrt{2}x + 5)$.
So the equation becomes: $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$.
Step 6: Factor out the common binomial term.
The common binomial term is $(\sqrt{2}x + 5)$.
$(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$.
Step 7: Set each factor to zero and solve for $x$.
Case 1: $\sqrt{2}x + 5 = 0$.
$\sqrt{2}x = -5$.
$x = -\frac{5}{\sqrt{2}}$.
To rationalise the denominator, multiply the numerator and denominator by $\sqrt{2}$:
$x = -\frac{5\sqrt{2}}{2}$.
Case 2: $x + \sqrt{2} = 0$.
$x = -\sqrt{2}$.
Therefore, the roots of the quadratic equation are $x = -\frac{5\sqrt{2}}{2}$ and $x = -\sqrt{2}$.
Question:
Represent the following situation in the form of a quadratic equation:
The product of two consecutive positive integers is 306. We need to find the integers.
Concept in a Minute:
Consecutive positive integers are integers that follow each other in order, with a difference of 1. For example, 5 and 6 are consecutive integers. A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0.
Explanation:
Let the first positive integer be represented by the variable ‘x’.
Since the integers are consecutive, the next positive integer will be ‘x + 1’.
The problem states that the product of these two consecutive positive integers is 306.
Therefore, we can write the equation:
x * (x + 1) = 306
To represent this in the form of a quadratic equation, we need to expand the expression and set it equal to zero.
x * x + x * 1 = 306
x² + x = 306
Now, move the constant term to the left side of the equation:
x² + x – 306 = 0
This is the quadratic equation representing the given situation. The question asks to represent the situation in the form of a quadratic equation, which we have done.
To find the integers (though not explicitly asked in the final representation part of the question, it is implied by “We need to find the integers”), we would solve this quadratic equation. We can factor it or use the quadratic formula.
Factoring: We need two numbers that multiply to -306 and add to 1. These numbers are 18 and -17.
So, (x + 18)(x – 17) = 0
This gives two possible values for x:
x + 18 = 0 => x = -18
x – 17 = 0 => x = 17
Since the question specifies “positive integers”, we discard the negative solution.
Therefore, the first positive integer (x) is 17.
The next consecutive positive integer (x + 1) is 17 + 1 = 18.
Let’s check the product: 17 * 18 = 306. This confirms our solution.
The quadratic equation representing the situation is x² + x – 306 = 0.
Question:
Check whether the following is the quadratic equation:
x3 – 4x2 – x + 1 = (x – 2)3
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree, meaning the highest power of the variable is 2. To check if an equation is quadratic, we need to simplify it and see if it can be expressed in the standard form ax^2 + bx + c = 0, where a, b, and c are constants and a is not equal to 0.
Explanation:
The given equation is x^3 – 4x^2 – x + 1 = (x – 2)^3.
To determine if this is a quadratic equation, we need to expand the right side of the equation and simplify the entire expression.
First, let’s expand (x – 2)^3 using the binomial expansion formula (a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3.
Here, a = x and b = 2.
So, (x – 2)^3 = x^3 – 3(x^2)(2) + 3(x)(2^2) – 2^3
(x – 2)^3 = x^3 – 6x^2 + 12x – 8.
Now, substitute this back into the original equation:
x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8.
Next, let’s move all the terms to one side of the equation to simplify it. We can subtract x^3 from both sides.
-4x^2 – x + 1 = -6x^2 + 12x – 8.
Now, let’s move all the terms from the right side to the left side. Add 6x^2 to both sides:
-4x^2 + 6x^2 – x + 1 = 12x – 8
2x^2 – x + 1 = 12x – 8.
Subtract 12x from both sides:
2x^2 – x – 12x + 1 = -8
2x^2 – 13x + 1 = -8.
Add 8 to both sides:
2x^2 – 13x + 1 + 8 = 0
2x^2 – 13x + 9 = 0.
The simplified equation is 2x^2 – 13x + 9 = 0.
This equation is in the standard form of a quadratic equation, ax^2 + bx + c = 0, where a = 2, b = -13, and c = 9. Since the highest power of x is 2 and the coefficient of x^2 (which is 2) is not zero, this is a quadratic equation.
Final Answer Check:
The highest power of the variable x in the simplified equation is 2. Therefore, the equation is quadratic.
Question:
Find two numbers whose sum is 27 and product is 182.
Concept in a Minute:
The problem requires setting up and solving a quadratic equation. We will use the concept that if we have two numbers, say x and y, their sum is x + y and their product is x * y. We can represent this in terms of a single variable by using the sum.
Explanation:
Let the two numbers be x and y.
According to the problem statement:
1. The sum of the two numbers is 27: x + y = 27
2. The product of the two numbers is 182: x * y = 182
From equation (1), we can express y in terms of x:
y = 27 – x
Now, substitute this expression for y into equation (2):
x * (27 – x) = 182
Expand the equation:
27x – x^2 = 182
Rearrange the terms to form a standard quadratic equation (ax^2 + bx + c = 0):
x^2 – 27x + 182 = 0
Now we need to solve this quadratic equation. We can do this by factoring, using the quadratic formula, or completing the square. Let’s try factoring. We need to find two numbers that multiply to 182 and add up to -27.
Let’s list factors of 182:
1 x 182
2 x 91
7 x 26
13 x 14
The pair 13 and 14 adds up to 27. So, -13 and -14 will add up to -27 and multiply to 182.
So, we can factor the quadratic equation as:
(x – 13)(x – 14) = 0
This gives us two possible values for x:
x – 13 = 0 => x = 13
x – 14 = 0 => x = 14
Case 1: If x = 13
Substitute x = 13 into the equation y = 27 – x:
y = 27 – 13
y = 14
Case 2: If x = 14
Substitute x = 14 into the equation y = 27 – x:
y = 27 – 14
y = 13
In both cases, the two numbers are 13 and 14.
Let’s check our answer:
Sum: 13 + 14 = 27 (Correct)
Product: 13 * 14 = 182 (Correct)
The two numbers are 13 and 14.
Question:
Check whether the following is the quadratic equation:
(2x – 1)(x – 3) = (x + 5)(x – 1)
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants and a is not equal to 0. To check if an equation is quadratic, we need to simplify it and see if it can be rearranged into this standard form, with the highest power of the variable being 2.
Explanation:
To determine if the given equation (2x – 1)(x – 3) = (x + 5)(x – 1) is a quadratic equation, we need to expand both sides of the equation and simplify it.
Step 1: Expand the left side of the equation.
(2x – 1)(x – 3) = 2x(x – 3) – 1(x – 3)
= 2x^2 – 6x – x + 3
= 2x^2 – 7x + 3
Step 2: Expand the right side of the equation.
(x + 5)(x – 1) = x(x – 1) + 5(x – 1)
= x^2 – x + 5x – 5
= x^2 + 4x – 5
Step 3: Set the expanded sides equal to each other.
2x^2 – 7x + 3 = x^2 + 4x – 5
Step 4: Rearrange the equation to bring all terms to one side, aiming for the standard quadratic form ax^2 + bx + c = 0.
Subtract x^2 from both sides:
(2x^2 – x^2) – 7x + 3 = 4x – 5
x^2 – 7x + 3 = 4x – 5
Subtract 4x from both sides:
x^2 – 7x – 4x + 3 = -5
x^2 – 11x + 3 = -5
Add 5 to both sides:
x^2 – 11x + 3 + 5 = 0
x^2 – 11x + 8 = 0
Step 5: Analyze the simplified equation.
The simplified equation is x^2 – 11x + 8 = 0.
This equation is in the form ax^2 + bx + c = 0, where a = 1, b = -11, and c = 8.
Since the highest power of the variable x is 2 and the coefficient ‘a’ (which is 1) is not zero, the equation is a quadratic equation.
Answer:
Yes, the given equation is a quadratic equation.
Question:
Check whether the following are the quadratic equation:
(x + 1)2 = 2(x – 3)
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term that is squared. The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. To check if an equation is quadratic, we need to simplify it and see if it can be expressed in this standard form with the highest power of the variable being 2.
Explanation:
We are given the equation (x + 1)² = 2(x – 3).
To check if this is a quadratic equation, we need to expand both sides and simplify it to see if it matches the standard form of a quadratic equation.
Step 1: Expand the left side of the equation.
Using the algebraic identity (a + b)² = a² + 2ab + b², we have:
(x + 1)² = x² + 2(x)(1) + 1² = x² + 2x + 1
Step 2: Expand the right side of the equation.
Distribute the 2 to the terms inside the parenthesis:
2(x – 3) = 2x – 2(3) = 2x – 6
Step 3: Set the expanded sides equal to each other.
x² + 2x + 1 = 2x – 6
Step 4: Rearrange the equation to bring all terms to one side, setting it equal to zero.
Subtract 2x from both sides:
x² + 2x – 2x + 1 = 2x – 2x – 6
x² + 1 = -6
Add 6 to both sides:
x² + 1 + 6 = -6 + 6
x² + 7 = 0
Step 5: Analyze the simplified equation.
The simplified equation is x² + 7 = 0.
This equation is in the form ax² + bx + c = 0, where a = 1, b = 0, and c = 7.
Since the highest power of the variable ‘x’ is 2, and the coefficient of x² (which is ‘a’) is not zero, this is a quadratic equation.
Answer: Yes, the given equation is a quadratic equation.
Question:
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Concept in a Minute:
The Pythagorean theorem relates the sides of a right triangle: a^2 + b^2 = c^2, where ‘a’ and ‘b’ are the lengths of the legs (base and altitude) and ‘c’ is the length of the hypotenuse. We will also use algebraic substitution to solve for unknown values.
Explanation:
Let the base of the right triangle be ‘x’ cm.
According to the question, the altitude of the right triangle is 7 cm less than its base, so the altitude is (x – 7) cm.
The hypotenuse is given as 13 cm.
Using the Pythagorean theorem:
(base)^2 + (altitude)^2 = (hypotenuse)^2
x^2 + (x – 7)^2 = 13^2
Expand the equation:
x^2 + (x^2 – 14x + 49) = 169
Combine like terms:
2x^2 – 14x + 49 = 169
Subtract 169 from both sides to set the equation to zero:
2x^2 – 14x + 49 – 169 = 0
2x^2 – 14x – 120 = 0
Divide the entire equation by 2 to simplify:
x^2 – 7x – 60 = 0
Now, we need to factor this quadratic equation. We are looking for two numbers that multiply to -60 and add up to -7. These numbers are -12 and 5.
So, we can rewrite the equation as:
(x – 12)(x + 5) = 0
This gives us two possible values for x:
x – 12 = 0 => x = 12
x + 5 = 0 => x = -5
Since the length of a side cannot be negative, we discard x = -5.
Therefore, the base of the right triangle is x = 12 cm.
Now, calculate the altitude:
Altitude = x – 7 = 12 – 7 = 5 cm.
So, the other two sides of the right triangle are 12 cm and 5 cm.
We can verify this using the Pythagorean theorem:
12^2 + 5^2 = 144 + 25 = 169
13^2 = 169
So, 12^2 + 5^2 = 13^2, which confirms our answer.
Question:
Check whether the following is the quadratic equation:
x2 + 3x + 1 = (x – 2)2
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is ax^2 + bx + c = 0, where ‘a’, ‘b’, and ‘c’ are constants and ‘a’ is not equal to 0. To check if an equation is quadratic, we need to simplify it and see if the highest power of the variable is 2 after simplification.
Explanation:
To check if the given equation x^2 + 3x + 1 = (x – 2)^2 is a quadratic equation, we need to simplify it and see if it can be expressed in the standard form of a quadratic equation.
Step 1: Expand the right side of the equation.
The right side is (x – 2)^2. Using the algebraic identity (a – b)^2 = a^2 – 2ab + b^2, we get:
(x – 2)^2 = x^2 – 2(x)(2) + 2^2
(x – 2)^2 = x^2 – 4x + 4
Step 2: Substitute the expanded form back into the original equation.
x^2 + 3x + 1 = x^2 – 4x + 4
Step 3: Rearrange the equation to bring all terms to one side, setting the equation to zero.
Subtract x^2 from both sides:
(x^2 – x^2) + 3x + 1 = -4x + 4
0 + 3x + 1 = -4x + 4
3x + 1 = -4x + 4
Add 4x to both sides:
3x + 4x + 1 = 4
7x + 1 = 4
Subtract 4 from both sides:
7x + 1 – 4 = 0
7x – 3 = 0
Step 4: Analyze the simplified equation.
The simplified equation is 7x – 3 = 0.
In this equation, the highest power of the variable ‘x’ is 1. This is a linear equation, not a quadratic equation. For an equation to be quadratic, the highest power of the variable must be 2.
Therefore, x^2 + 3x + 1 = (x – 2)^2 is not a quadratic equation.
Question:
Check whether the following is the quadratic equation:
(x – 2)(x + 1) = (x – 1)(x + 3)
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree. This means that the highest power of the variable in the equation is 2. A general form of a quadratic equation is ax^2 + bx + c = 0, where ‘a’ is not equal to 0. To check if an equation is quadratic, we need to simplify it and see if it can be expressed in this standard form.
Explanation:
We are given the equation (x – 2)(x + 1) = (x – 1)(x + 3).
To check if this is a quadratic equation, we need to expand both sides of the equation and simplify it.
Expand the left side of the equation:
(x – 2)(x + 1) = x(x + 1) – 2(x + 1)
= x^2 + x – 2x – 2
= x^2 – x – 2
Expand the right side of the equation:
(x – 1)(x + 3) = x(x + 3) – 1(x + 3)
= x^2 + 3x – x – 3
= x^2 + 2x – 3
Now, set the expanded left side equal to the expanded right side:
x^2 – x – 2 = x^2 + 2x – 3
To simplify the equation, we move all terms to one side. Let’s move all terms from the right side to the left side:
x^2 – x – 2 – (x^2 + 2x – 3) = 0
x^2 – x – 2 – x^2 – 2x + 3 = 0
Combine like terms:
(x^2 – x^2) + (-x – 2x) + (-2 + 3) = 0
0 + (-3x) + (1) = 0
-3x + 1 = 0
This simplified equation is -3x + 1 = 0.
The highest power of the variable ‘x’ in this equation is 1.
A quadratic equation must have the highest power of the variable as 2.
Since the highest power of x is 1, this equation is a linear equation, not a quadratic equation.
Therefore, the given equation is not a quadratic equation.
Question:
Represent the following situation in the form of a quadratic equation:
Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Concept in a Minute:
Quadratic Equation Formation: Translate a word problem involving relationships between ages and their products into a quadratic equation.
Algebraic Manipulation: Set up an equation by defining variables for unknown ages and using the given information to express other unknown quantities in terms of these variables.
Solving Quadratic Equations: Once the quadratic equation is formed, it can be solved using methods like factorization, completing the square, or the quadratic formula to find the value of the variable.
Explanation:
Let Rohan’s present age be ‘x’ years.
Since Rohan’s mother is 26 years older than him, her present age will be (x + 26) years.
In 3 years from now:
Rohan’s age will be (x + 3) years.
Rohan’s mother’s age will be (x + 26 + 3) = (x + 29) years.
According to the problem, the product of their ages 3 years from now will be 360.
So, we can write the equation:
(x + 3)(x + 29) = 360
Now, expand the left side of the equation:
x * x + x * 29 + 3 * x + 3 * 29 = 360
x^2 + 29x + 3x + 87 = 360
x^2 + 32x + 87 = 360
To form a quadratic equation in the standard form (ax^2 + bx + c = 0), move all terms to one side:
x^2 + 32x + 87 – 360 = 0
x^2 + 32x – 273 = 0
This is the quadratic equation that represents the given situation. The question asks to represent the situation in the form of a quadratic equation.
The final answer is $\boxed{x^2 + 32x – 273 = 0}$.
Question:
Find the roots of the following quadratic equation by factorisation:
2x2 + x – 6 = 0
Concept in a Minute:
Quadratic equations are equations of the form ax^2 + bx + c = 0, where a, b, and c are constants and a is not equal to 0. The roots of a quadratic equation are the values of x that satisfy the equation. Factorisation is a method to find the roots by expressing the quadratic expression as a product of two linear factors. This involves splitting the middle term (bx) into two terms such that their sum is equal to bx and their product is equal to ac.
Explanation:
The given quadratic equation is 2x^2 + x – 6 = 0.
To solve this by factorisation, we need to find two numbers that multiply to (2 * -6) = -12 and add up to the coefficient of x, which is 1.
These two numbers are 4 and -3.
We can rewrite the middle term (x) as 4x – 3x.
So, the equation becomes:
2x^2 + 4x – 3x – 6 = 0
Now, we group the terms and factor out common factors:
(2x^2 + 4x) + (– 3x – 6) = 0
2x(x + 2) – 3(x + 2) = 0
Now, we can factor out the common binomial factor (x + 2):
(2x – 3)(x + 2) = 0
For the product of two factors to be zero, at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for x:
Case 1: 2x – 3 = 0
Add 3 to both sides:
2x = 3
Divide by 2:
x = 3/2
Case 2: x + 2 = 0
Subtract 2 from both sides:
x = -2
Therefore, the roots of the quadratic equation 2x^2 + x – 6 = 0 are x = 3/2 and x = -2.
Question:
Find the roots of the following quadratic equation by factorisation:
$2𝑥2–𝑥+18=0$
Concept in a Minute:
To find the roots of a quadratic equation by factorization, we need to express the quadratic expression as a product of two linear factors. This is usually achieved by splitting the middle term. The roots are the values of the variable that make these factors equal to zero. The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Explanation:
The given quadratic equation is $2x^2 – x + 1/8 = 0$.
First, let’s clear the fraction by multiplying the entire equation by 8:
$8 * (2x^2 – x + 1/8) = 8 * 0$
$16x^2 – 8x + 1 = 0$
Now, we need to factorize this quadratic equation. We are looking for two numbers that multiply to $a*c$ (which is $16 * 1 = 16$) and add up to $b$ (which is $-8$).
The two numbers are -4 and -4, because $(-4) * (-4) = 16$ and $(-4) + (-4) = -8$.
Now, we split the middle term (-8x) using these two numbers:
$16x^2 – 4x – 4x + 1 = 0$
Next, we group the terms and factor out common factors:
$(16x^2 – 4x) + (– 4x + 1) = 0$
$4x(4x – 1) – 1(4x – 1) = 0$
Now, we factor out the common binomial factor $(4x – 1)$:
$(4x – 1)(4x – 1) = 0$
To find the roots, we set each factor equal to zero:
$4x – 1 = 0$
$4x = 1$
$x = 1/4$
Since both factors are the same, we have a repeated root.
Therefore, the roots of the quadratic equation $2x^2 – x + 1/8 = 0$ are $x = 1/4$ and $x = 1/4$.
Question:
Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
$3𝑥2−4\sqrt{3𝑥+4=0}$
Concept in a Minute:
The nature of roots of a quadratic equation $ax^2 + bx + c = 0$ is determined by its discriminant, $\Delta = b^2 – 4ac$.
– If $\Delta > 0$, the roots are real and distinct.
– If $\Delta = 0$, the roots are real and equal.
– If $\Delta < 0$, the roots are not real (complex).
If real roots exist, they can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.
Explanation:
The given quadratic equation is $3x^2 – 4\sqrt{3}x + 4 = 0$.
This equation is in the standard form $ax^2 + bx + c = 0$, where:
$a = 3$
$b = -4\sqrt{3}$
$c = 4$
First, we need to find the discriminant ($\Delta$) to determine the nature of the roots.
$\Delta = b^2 – 4ac$
Substitute the values of a, b, and c:
$\Delta = (-4\sqrt{3})^2 – 4(3)(4)$
$\Delta = (16 \times 3) – 48$
$\Delta = 48 – 48$
$\Delta = 0$
Since the discriminant $\Delta = 0$, the quadratic equation has real and equal roots.
Now, we find the roots using the quadratic formula:
$x = \frac{-b \pm \sqrt{\Delta}}{2a}$
Substitute the values of a, b, and $\Delta$:
$x = \frac{-(-4\sqrt{3}) \pm \sqrt{0}}{2(3)}$
$x = \frac{4\sqrt{3} \pm 0}{6}$
$x = \frac{4\sqrt{3}}{6}$
Simplify the expression:
$x = \frac{2\sqrt{3}}{3}$
Therefore, the nature of the roots is real and equal, and the roots are $x = \frac{2\sqrt{3}}{3}$.
Question:
Check whether the following is the quadratic equation:
(x + 2)3 = 2x(x2 – 1)
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree. This means the highest power of the variable in the equation must be 2. To check if an equation is quadratic, we need to expand and simplify it, and then determine the highest power of the variable. If the highest power is 2, it’s a quadratic equation; otherwise, it’s not.
Explanation:
The given equation is (x + 2)³ = 2x(x² – 1).
To check if this is a quadratic equation, we need to expand both sides and simplify.
First, expand the left side using the binomial expansion formula (a + b)³ = a³ + 3a²b + 3ab² + b³:
(x + 2)³ = x³ + 3(x²)(2) + 3(x)(2²) + 2³
(x + 2)³ = x³ + 6x² + 12x + 8
Next, expand the right side:
2x(x² – 1) = 2x * x² – 2x * 1
2x(x² – 1) = 2x³ – 2x
Now, set the expanded left side equal to the expanded right side:
x³ + 6x² + 12x + 8 = 2x³ – 2x
To simplify and determine the highest power, move all terms to one side of the equation. Let’s move all terms to the right side:
0 = 2x³ – x³ – 6x² – 2x – 12x – 8
0 = x³ – 6x² – 14x – 8
The simplified equation is x³ – 6x² – 14x – 8 = 0.
The highest power of the variable ‘x’ in this equation is 3.
Since the highest power is 3, this is a cubic equation, not a quadratic equation.
Therefore, the given equation is not a quadratic equation.
Question:
Check whether the following is the quadratic equation:
(x – 3)(2x + 1) = x(x + 5)
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree, meaning it can be written in the standard form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. To check if an equation is quadratic, we need to simplify it and see if the highest power of the variable is 2.
Explanation:
To determine if the given equation is quadratic, we need to expand both sides and simplify it into the standard form ax² + bx + c = 0.
Step 1: Expand the left side of the equation.
(x – 3)(2x + 1) = x(2x) + x(1) – 3(2x) – 3(1)
= 2x² + x – 6x – 3
= 2x² – 5x – 3
Step 2: Expand the right side of the equation.
x(x + 5) = x(x) + x(5)
= x² + 5x
Step 3: Set the expanded left side equal to the expanded right side.
2x² – 5x – 3 = x² + 5x
Step 4: Rearrange the terms to bring all terms to one side to get the standard form ax² + bx + c = 0.
Subtract x² from both sides:
2x² – x² – 5x – 3 = 5x
x² – 5x – 3 = 5x
Subtract 5x from both sides:
x² – 5x – 5x – 3 = 0
x² – 10x – 3 = 0
Step 5: Check the highest power of the variable.
The simplified equation is x² – 10x – 3 = 0. The highest power of x is 2.
Conclusion:
Since the simplified equation is in the form ax² + bx + c = 0, where a = 1, b = -10, and c = -3 (and a ≠ 0), it is a quadratic equation.
Question:
Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
2x2 – 6x + 3 = 0
Concept in a Minute:
The nature of the roots of a quadratic equation ax^2 + bx + c = 0 is determined by its discriminant, D = b^2 – 4ac.
– If D > 0, the roots are real and distinct.
– If D = 0, the roots are real and equal.
– If D < 0, the roots are not real (complex).
If real roots exist (D >= 0), they can be found using the quadratic formula: x = [-b ± sqrt(D)] / 2a.
Explanation:
We are given the quadratic equation 2x^2 – 6x + 3 = 0.
Here, a = 2, b = -6, and c = 3.
First, we find the discriminant (D) to determine the nature of the roots:
D = b^2 – 4ac
D = (-6)^2 – 4 * (2) * (3)
D = 36 – 24
D = 12
Since D = 12, which is greater than 0 (D > 0), the roots of the equation are real and distinct.
Now, we find the real roots using the quadratic formula:
x = [-b ± sqrt(D)] / 2a
x = [-(-6) ± sqrt(12)] / (2 * 2)
x = [6 ± sqrt(12)] / 4
We can simplify sqrt(12) as sqrt(4 * 3) = 2 * sqrt(3).
So, x = [6 ± 2 * sqrt(3)] / 4
Now, we can split this into two roots:
x1 = (6 + 2 * sqrt(3)) / 4
x1 = 2 * (3 + sqrt(3)) / 4
x1 = (3 + sqrt(3)) / 2
x2 = (6 – 2 * sqrt(3)) / 4
x2 = 2 * (3 – sqrt(3)) / 4
x2 = (3 – sqrt(3)) / 2
Therefore, the nature of the roots is real and distinct, and the roots are (3 + sqrt(3))/2 and (3 – sqrt(3))/2.
Question:
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Concept in a Minute:
Area of a rectangle, algebraic equations, solving quadratic equations.
Explanation:
Let the breadth of the rectangular mango grove be ‘b’ meters.
According to the question, the length of the mango grove is twice its breadth.
So, the length ‘l’ = 2b meters.
The area of a rectangle is given by the formula: Area = length × breadth.
We are given that the area of the mango grove is 800 m².
Therefore, we can write the equation:
l × b = 800
Substitute the expression for length (l = 2b) into the area equation:
(2b) × b = 800
Simplify the equation:
2b² = 800
Now, we need to solve for ‘b’. Divide both sides of the equation by 2:
b² = 800 / 2
b² = 400
To find the value of ‘b’, take the square root of both sides:
b = √400
Since breadth must be a positive value, we consider the positive square root:
b = 20 meters
Now that we have the breadth, we can find the length using the relationship l = 2b:
l = 2 × 20
l = 40 meters
To verify our answer, we can calculate the area with these dimensions:
Area = length × breadth = 40 m × 20 m = 800 m²
This matches the given area.
So, it is possible to design such a rectangular mango grove.
The length of the mango grove is 40 meters, and its breadth is 20 meters.
Question:
Find the roots of the following quadratic equation by factorisation:
100x2 – 20x + 1 = 0
Concept in a Minute:
To find the roots of a quadratic equation by factorization, we need to express the quadratic equation in the form ax^2 + bx + c = 0 and then find two numbers whose product is ac and whose sum is b. These two numbers are then used to split the middle term, allowing us to factor the quadratic expression into two linear factors. Setting each factor to zero will give us the roots of the equation.
Explanation:
The given quadratic equation is 100x^2 – 20x + 1 = 0.
We need to find two numbers whose product is (100 * 1) = 100 and whose sum is -20.
These two numbers are -10 and -10, because (-10) * (-10) = 100 and (-10) + (-10) = -20.
Now, we split the middle term (-20x) using these two numbers:
100x^2 – 10x – 10x + 1 = 0
Next, we group the terms and factor out the common factors:
(100x^2 – 10x) + (-10x + 1) = 0
10x(10x – 1) – 1(10x – 1) = 0
Now, we can factor out the common binomial factor (10x – 1):
(10x – 1)(10x – 1) = 0
To find the roots, we set each factor equal to zero:
10x – 1 = 0
10x = 1
x = 1/10
Since both factors are the same, the roots are repeated.
Therefore, the roots of the quadratic equation 100x^2 – 20x + 1 = 0 are x = 1/10 and x = 1/10.
Question:
Check whether the following is the quadratic equation:
x2 – 2x = (-2)(3 – x)
Concept in a Minute:
A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants and a is not equal to 0. To check if an equation is quadratic, we need to simplify it and see if the highest power of the variable is 2 and if the coefficient of the x^2 term is non-zero.
Explanation:
The given equation is:
x^2 – 2x = (-2)(3 – x)
First, we simplify the right side of the equation by distributing the -2:
(-2)(3 – x) = (-2) * 3 + (-2) * (-x)
= -6 + 2x
Now, substitute this back into the original equation:
x^2 – 2x = -6 + 2x
Next, we want to rearrange the equation to the standard form of a quadratic equation (ax^2 + bx + c = 0) by moving all terms to one side. Subtract 2x from both sides:
x^2 – 2x – 2x = -6 + 2x – 2x
x^2 – 4x = -6
Now, add 6 to both sides:
x^2 – 4x + 6 = -6 + 6
x^2 – 4x + 6 = 0
This equation is in the form ax^2 + bx + c = 0, where a = 1, b = -4, and c = 6. Since the highest power of x is 2 and the coefficient of x^2 (which is a=1) is not zero, the given equation is indeed a quadratic equation.
Therefore, the answer is Yes.
Question:
Represent the following situation in the form of a quadratic equation:
The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Concept in a Minute:
Understanding and forming quadratic equations from word problems. Identifying variables, translating word phrases into algebraic expressions, and using given information (like area formula) to set up an equation.
Explanation:
Let the breadth of the rectangular plot be ‘x’ metres.
According to the problem, the length of the plot is one more than twice its breadth.
So, the length of the plot can be represented as (2x + 1) metres.
The area of a rectangle is given by the formula: Area = Length × Breadth.
We are given that the area of the rectangular plot is 528 m².
Substituting the expressions for length and breadth into the area formula:
(2x + 1) × x = 528
Expanding the equation:
2x² + x = 528
To represent this in the standard form of a quadratic equation (ax² + bx + c = 0), we move the constant term to the left side:
2x² + x – 528 = 0
This is the quadratic equation representing the given situation. To find the length and breadth, we would solve this equation for ‘x’ using methods like factorization, completing the square, or the quadratic formula.
Question:
Find the values of k for the following quadratic equation, so that they have two equal roots.
2x2 + kx + 3 = 0
Concept in a Minute:
A quadratic equation of the form ax^2 + bx + c = 0 has two equal roots if and only if its discriminant (D) is equal to zero. The discriminant is given by the formula D = b^2 – 4ac.
Explanation:
The given quadratic equation is 2x^2 + kx + 3 = 0.
We need to find the values of k for which this equation has two equal roots.
For a quadratic equation to have two equal roots, the discriminant must be zero.
In this equation, we have:
a = 2
b = k
c = 3
The discriminant D is calculated as:
D = b^2 – 4ac
Setting the discriminant to zero for equal roots:
D = 0
k^2 – 4(2)(3) = 0
k^2 – 24 = 0
Now, we solve for k:
k^2 = 24
k = ±√24
To simplify √24, we find the prime factorization of 24:
24 = 2 × 2 × 2 × 3 = 4 × 6
So, √24 = √(4 × 6) = √4 × √6 = 2√6
Therefore, the values of k are:
k = 2√6 or k = -2√6
Final Answer: The final answer is $\boxed{k = \pm 2\sqrt{6}}$
Question:
Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
2x2 – 3x + 5 = 0
Concept in a Minute:
The nature of the roots of a quadratic equation ax^2 + bx + c = 0 is determined by the discriminant, delta, which is calculated as delta = b^2 – 4ac.
If delta > 0, the roots are real and distinct.
If delta = 0, the roots are real and equal.
If delta < 0, the roots are not real (complex conjugates).
If real roots exist, they can be found using the quadratic formula: x = (-b ± sqrt(delta)) / 2a.
Explanation:
The given quadratic equation is 2x^2 – 3x + 5 = 0.
Here, a = 2, b = -3, and c = 5.
First, calculate the discriminant (delta):
delta = b^2 – 4ac
delta = (-3)^2 – 4 * (2) * (5)
delta = 9 – 40
delta = -31
Since delta (-31) is less than 0, the roots of the quadratic equation are not real. They are complex conjugates. Therefore, there are no real roots to find.
Next Chapter: Real Numbers
Refer Quadratic Equations Notes
Practice Quadratic Equations Extra Questions
Conquer Maths & Science – with LearnTheta’s AI-Practice!
✅ All Topics at One Place
🤖 Adaptive Question Practice
📊 Progress and Insights