NCERT Class 10 Maths Solutions: Polynomials

Question:

To find the zeroes of a quadratic polynomial, we set the polynomial equal to zero and solve for the variable. This can be done by factoring the quadratic expression. The relationship between the zeroes and coefficients of a quadratic polynomial ax² + bx + c are: Sum of zeroes = -b/a and Product of zeroes = c/a.


The given quadratic polynomial is x² – 2x – 8.

Step 1: Find the zeroes of the polynomial.
To find the zeroes, we set the polynomial equal to zero:
x² – 2x – 8 = 0

We can factor this quadratic equation. We need to find two numbers that multiply to -8 and add up to -2. These numbers are -4 and +2.
So, we can rewrite the equation as:
(x – 4)(x + 2) = 0

Now, we set each factor to zero to find the values of x:
x – 4 = 0 => x = 4
x + 2 = 0 => x = -2

Therefore, the zeroes of the quadratic polynomial x² – 2x – 8 are 4 and -2.

Step 2: Verify the relationship between the zeroes and the coefficients.
The general form of a quadratic polynomial is ax² + bx + c.
In our polynomial, x² – 2x – 8, we have:
a = 1
b = -2
c = -8

Let the zeroes be α and β. From Step 1, we found α = 4 and β = -2.

Verification of the sum of zeroes:
Sum of zeroes = α + β
Sum of zeroes = 4 + (-2) = 2

According to the relationship, Sum of zeroes = -b/a
Sum of zeroes = -(-2)/1 = 2

Since 2 = 2, the relationship for the sum of zeroes is verified.

Verification of the product of zeroes:
Product of zeroes = α * β
Product of zeroes = 4 * (-2) = -8

According to the relationship, Product of zeroes = c/a
Product of zeroes = -8/1 = -8

Since -8 = -8, the relationship for the product of zeroes is verified.

Thus, the zeroes of the quadratic polynomial x² – 2x – 8 are 4 and -2, and the relationships between the zeroes and coefficients are verified.

Question:

To find the zeroes of a quadratic polynomial, we set the polynomial equal to zero and solve for the variable. The zeroes are the values of the variable that make the polynomial equal to zero. For a quadratic polynomial of the form ax^2 + bx + c, the sum of the zeroes (alpha + beta) is equal to -b/a, and the product of the zeroes (alpha * beta) is equal to c/a.


The given quadratic polynomial is 4u^2 + 8u.
To find the zeroes, we set the polynomial equal to zero:
4u^2 + 8u = 0

We can factor out the common term, which is 4u:
4u(u + 2) = 0

For the product of two terms to be zero, at least one of the terms must be zero. Therefore, we have two possibilities:
1. 4u = 0
Dividing both sides by 4, we get:
u = 0

2. u + 2 = 0
Subtracting 2 from both sides, we get:
u = -2

So, the zeroes of the quadratic polynomial 4u^2 + 8u are 0 and -2.

Now, we need to verify the relationship between the zeroes and the coefficients.
The given polynomial is 4u^2 + 8u. This can be compared to the standard quadratic form ax^2 + bx + c.
Here, a = 4, b = 8, and c = 0 (since there is no constant term).

Let the zeroes be alpha = 0 and beta = -2.

Verification of the sum of zeroes:
Sum of zeroes = alpha + beta = 0 + (-2) = -2
According to the relationship, the sum of zeroes = -b/a = -8/4 = -2.
The sum of zeroes matches the calculated value (-2 = -2).

Verification of the product of zeroes:
Product of zeroes = alpha * beta = 0 * (-2) = 0
According to the relationship, the product of zeroes = c/a = 0/4 = 0.
The product of zeroes matches the calculated value (0 = 0).

Thus, the relationship between the zeroes and the coefficients is verified.

Question:

To find the zeroes of a quadratic polynomial, we set the polynomial equal to zero and solve for the variable. The zeroes are the values of the variable that make the polynomial equal to zero. For a quadratic polynomial of the form ax^2 + bx + c, if alpha and beta are the zeroes, then the sum of the zeroes (alpha + beta) is equal to -b/a, and the product of the zeroes (alpha * beta) is equal to c/a. In this question, the polynomial is in the form at^2 + c, which is a special case where b = 0.


To find the zeroes of the quadratic polynomial t^2 – 15, we set it equal to zero:
t^2 – 15 = 0

Add 15 to both sides of the equation:
t^2 = 15

Take the square root of both sides to solve for t:
t = ±√15

So, the zeroes of the quadratic polynomial t^2 – 15 are √15 and -√15.

Now, let’s verify the relationship between the zeroes and the coefficients.
The given polynomial is t^2 – 15. We can write this in the standard form of a quadratic polynomial as 1*t^2 + 0*t – 15.
Here, a = 1, b = 0, and c = -15.

Let the zeroes be α = √15 and β = -√15.

Sum of the zeroes:
α + β = √15 + (-√15) = √15 – √15 = 0

According to the relationship, the sum of the zeroes should be -b/a:
-b/a = -(0)/1 = 0

The calculated sum of zeroes (0) matches the expected sum of zeroes (-b/a = 0).

Product of the zeroes:
α * β = (√15) * (-√15) = -(√15)^2 = -15

According to the relationship, the product of the zeroes should be c/a:
c/a = (-15)/1 = -15

The calculated product of zeroes (-15) matches the expected product of zeroes (c/a = -15).

Thus, the relationship between the zeroes and the coefficients is verified.

Question:

The relationship between the zeroes of a quadratic polynomial and its coefficients. For a quadratic polynomial $ax^2 + bx + c$, if $\alpha$ and $\beta$ are its zeroes, then the sum of the zeroes is $\alpha + \beta = -\frac{b}{a}$ and the product of the zeroes is $\alpha \beta = \frac{c}{a}$. A general form of a quadratic polynomial with zeroes $\alpha$ and $\beta$ can be written as $k(x^2 – (\alpha + \beta)x + \alpha \beta)$, where $k$ is any non-zero constant.


Let the quadratic polynomial be $P(x) = ax^2 + bx + c$.
Let the zeroes of the polynomial be $\alpha$ and $\beta$.
We are given that the sum of the zeroes is $-14$, so $\alpha + \beta = -14$.
We are also given that the product of the zeroes is $14$, so $\alpha \beta = 14$.

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes using the formula:
$P(x) = k(x^2 – (\text{sum of zeroes})x + \text{product of zeroes})$
where $k$ is a non-zero constant.

Substitute the given values of the sum and product of zeroes into the formula:
$P(x) = k(x^2 – (-14)x + 14)$
$P(x) = k(x^2 + 14x + 14)$

We can choose any non-zero value for $k$. The simplest choice is $k=1$.
So, the quadratic polynomial is:
$P(x) = x^2 + 14x + 14$

Thus, a quadratic polynomial with the sum of its zeroes as $-14$ and the product of its zeroes as $14$ is $x^2 + 14x + 14$.

Question:

The relationship between the zeroes of a quadratic polynomial and its coefficients. For a quadratic polynomial $ax^2 + bx + c$, if $\alpha$ and $\beta$ are its zeroes, then the sum of the zeroes is $\alpha + \beta = -b/a$ and the product of the zeroes is $\alpha\beta = c/a$. A general form of a quadratic polynomial with zeroes $\alpha$ and $\beta$ can be written as $k(x^2 – (\alpha + \beta)x + \alpha\beta)$, where $k$ is a non-zero constant.


We are given that the sum of the zeroes is 1 and the product of the zeroes is 1.
Let the quadratic polynomial be $P(x) = ax^2 + bx + c$.
Let the zeroes of the polynomial be $\alpha$ and $\beta$.
According to the problem statement, we have:
Sum of zeroes, $\alpha + \beta = 1$
Product of zeroes, $\alpha\beta = 1$

We know that a quadratic polynomial can be expressed in terms of the sum and product of its zeroes as:
$P(x) = k(x^2 – (\alpha + \beta)x + \alpha\beta)$
where $k$ is a non-zero constant.

Substitute the given values of the sum and product of the zeroes into this formula:
$P(x) = k(x^2 – (1)x + 1)$
$P(x) = k(x^2 – x + 1)$

We can choose any non-zero value for $k$. For simplicity, let’s take $k = 1$.
So, the quadratic polynomial is:
$P(x) = 1(x^2 – x + 1)$
$P(x) = x^2 – x + 1$

Thus, the quadratic polynomial is $x^2 – x + 1$.

Question:

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes. If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial $ax^2 + bx + c$, then the sum of the zeroes is $\alpha + \beta = -\frac{b}{a}$ and the product of the zeroes is $\alpha \beta = \frac{c}{a}$. A general form of a quadratic polynomial with zeroes $\alpha$ and $\beta$ is given by $k(x^2 – (\alpha + \beta)x + \alpha \beta)$, where $k$ is any non-zero constant.


The question asks us to find a quadratic polynomial given the sum and product of its zeroes.
We are given that the sum of the zeroes is 4 and the product of the zeroes is 1.
Let the sum of the zeroes be $S$ and the product of the zeroes be $P$.
So, $S = 4$ and $P = 1$.
The general form of a quadratic polynomial with sum of zeroes $S$ and product of zeroes $P$ is $x^2 – Sx + P$.
Substituting the given values, we get:
$x^2 – (4)x + (1)$
$x^2 – 4x + 1$
This is a quadratic polynomial. We can also multiply this by any non-zero constant $k$ to get other possible quadratic polynomials, such as $k(x^2 – 4x + 1)$. However, the simplest form is usually preferred, which is when $k=1$.

Thus, the quadratic polynomial is $x^2 – 4x + 1$.

Final Answer: The final answer is $\boxed{x^2 – 4x + 1}$

Question:

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes. If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial, then the polynomial can be written as $k(x^2 – (\alpha + \beta)x + \alpha\beta)$, where $k$ is any non-zero constant.


Let the given numbers be the sum of the zeroes and the product of the zeroes respectively.
Given:
Sum of zeroes ($\alpha + \beta$) = 0
Product of zeroes ($\alpha\beta$) = $\sqrt{5}$

The general form of a quadratic polynomial with zeroes $\alpha$ and $\beta$ is given by:
$P(x) = k(x^2 – (\alpha + \beta)x + \alpha\beta)$

Substitute the given values of the sum and product of zeroes into the formula:
$P(x) = k(x^2 – (0)x + \sqrt{5})$
$P(x) = k(x^2 + \sqrt{5})$

We can choose any non-zero value for $k$. The simplest choice is $k=1$.
Therefore, the quadratic polynomial is:
$P(x) = x^2 + \sqrt{5}$

Thus, a quadratic polynomial with the sum of its zeroes as 0 and the product of its zeroes as $\sqrt{5}$ is $x^2 + \sqrt{5}$.

Question:

A quadratic polynomial can be represented as $ax^2 + bx + c$. For a quadratic polynomial, if the zeroes are $\alpha$ and $\beta$, then the sum of the zeroes is $\alpha + \beta = -b/a$ and the product of the zeroes is $\alpha \beta = c/a$. A common form of a quadratic polynomial is $x^2 – (\text{sum of zeroes})x + (\text{product of zeroes})$.


Let the quadratic polynomial be $P(x)$.
Let the zeroes of the polynomial be $\alpha$ and $\beta$.
We are given the sum of the zeroes, $\alpha + \beta = 14$.
We are also given the product of the zeroes, $\alpha \beta = -1$.

A quadratic polynomial can be constructed using the sum and product of its zeroes with the formula:
$P(x) = k[x^2 – (\text{sum of zeroes})x + (\text{product of zeroes})]$
where $k$ is any non-zero constant.

For simplicity, we can choose $k=1$.
So, the quadratic polynomial will be:
$P(x) = x^2 – (14)x + (-1)$
$P(x) = x^2 – 14x – 1$

Therefore, the quadratic polynomial is $x^2 – 14x – 1$.