Class 10 Maths: Quadratic Equations – Extra Questions with Answers

Ans: A

Solution:
We can factor the quadratic as $(x-3)(x-1) = 0$. Thus, $x-3=0$ or $x-1=0$. This yields $x=3$ or $x=1$.
Answer: $x=1, 3$

Ans: B

Solution:
The sum of the roots of a quadratic equation $ax^2 + bx + c = 0$ is given by $-b/a$. In the equation $x^2 – 7x + 12 = 0$, $a = 1$ and $b = -7$. Therefore, the sum of the roots is $-(-7)/1 = 7$.
Answer: 7

Ans: A

Solution:
For the quadratic equation $ax^2 + bx + c = 0$ to have real roots, the discriminant $b^2 – 4ac$ must be greater than or equal to 0.
In this case, $a = 1$, $b = k$, and $c = -4$.
So, the discriminant is $k^2 – 4(1)(-4) = k^2 + 16$.
We need $k^2 + 16 \ge 0$. Since $k^2$ is always non-negative, $k^2 + 16$ will always be greater than or equal to 16, which is always greater than or equal to 0.
Therefore, the condition $k^2 + 16 \ge 0$ is true for all real values of $k$.

Answer: $(-\infty, \infty)$

Ans: C

Solution:
$(3x – 2)(3x – 2) = 0$
$3x – 2 = 0$
$3x = 2$
$x = \frac{2}{3}$
Answer: $\frac{2}{3}, \frac{2}{3}$

Ans: B

Solution: Let the base be $b$ meters and the altitude be $h$ meters. We are given that $h = b+3$. The area of the triangle is given by $\frac{1}{2}bh = 170$. Substituting $h = b+3$ into the area formula, we get $\frac{1}{2}b(b+3) = 170$, so $b(b+3) = 340$. Expanding, we have $b^2 + 3b – 340 = 0$. Factoring, we have $(b-17)(b+20)=0$. Since the base cannot be negative, $b=17$. Then $h = b+3 = 17+3 = 20$.
Answer: Base: 17 m, Altitude: 20 m

Ans: A

Solution:
Let $u = \frac{x}{x+2}$. Then the equation becomes $u^2 – 4u + 3 = 0$. Factoring, we get $(u-3)(u-1)=0$, so $u=3$ or $u=1$.
If $u=3$, then $\frac{x}{x+2}=3$, so $x = 3x+6$, which implies $2x=-6$, so $x=-3$.
If $u=1$, then $\frac{x}{x+2}=1$, so $x = x+2$, which has no solution.
Thus $x=-3$.

Answer: -3

Ans: D

Solution:
The discriminant is $b^2 – 4ac = 6^2 – 4(2\sqrt{2})(\sqrt{2}) = 36 – 16 = 20$.
Answer: 20

Ans: C

Solution:
For real and distinct roots, the discriminant $D = k^2 – 4ac > 0$.
Here, $a=2, b=k, c=8$.
So, $k^2 – 4(2)(8) > 0$
$k^2 – 64 > 0$
$k^2 > 64$
$k > 8$ or $k < -8$.
Answer: $k<-8$ or $k>8$

Ans: A

Solution:
Let the quadratic equation be $ax^2 + bx + c = 0$.
We are given that $x_1x_2 = \frac{c}{a} = 9$.
Also, $\frac{x_1}{x_1 – 2} + \frac{x_2}{x_2 – 2} = 2$.
$\frac{x_1(x_2 – 2) + x_2(x_1 – 2)}{(x_1 – 2)(x_2 – 2)} = 2$
$\frac{2x_1x_2 – 2(x_1 + x_2)}{x_1x_2 – 2(x_1 + x_2) + 4} = 2$
$2x_1x_2 – 2(x_1 + x_2) = 2(x_1x_2 – 2(x_1 + x_2) + 4)$
$2(9) – 2(x_1 + x_2) = 2(9 – 2(x_1 + x_2) + 4)$
$18 – 2(x_1 + x_2) = 18 – 4(x_1 + x_2) + 8$
$2(x_1 + x_2) = 8$
$x_1 + x_2 = 4$
Using Vieta’s formulas, $x_1 + x_2 = -\frac{b}{a} = 4$ and $x_1x_2 = \frac{c}{a} = 9$.
Let $a = 1$. Then $b = -4$ and $c = 9$.
The quadratic equation is $x^2 – 4x + 9 = 0$.

Answer: $x^2 – 4x + 9 = 0$

Ans: D

Solution:
Let the annual percentage increase be $r$.
Let the revenue at the beginning of the first year be $R$.
After the first year, the revenue becomes $R(1+r)$.
After the second year, the revenue becomes $R(1+r)(1+r) = R(1+r)^2$.
We are given that the revenue tripled over the last two years, so
$R(1+r)^2 = 3R$.
Dividing both sides by $R$, we have $(1+r)^2 = 3$.
Taking the square root of both sides, we have $1+r = \sqrt{3}$.
Therefore, $r = \sqrt{3}-1$.
To find the percentage increase, we have $r = \sqrt{3}-1 \approx 1.732 – 1 = 0.732$.
The annual percentage increase is $(\sqrt{3}-1) \times 100 \%$.
The value of the annual percentage increase is $(\sqrt{3}-1) \times 100\% \approx 73.2 \%$.

Answer: $(\sqrt{3}-1) \times 100\%$