10 Important Quadratic Equations Questions and Solutions for Class 10

10 Quadratic Equations Questions with Solution

Solutions

Question 1:

Solve the quadratic equation $x^2 - 5x + 6 = 0$ by factorization method.

Solution:

Given equation: $x^2 - 5x + 6 = 0$

To factorize, we look for two numbers whose sum is $-5$ and product is $6$. These numbers are $-2$ and $-3$.

$x^2 - 2x - 3x + 6 = 0$

$x(x - 2) - 3(x - 2) = 0$

$(x - 2)(x - 3) = 0$

Therefore, $x - 2 = 0$ or $x - 3 = 0$

$x = 2$ or $x = 3$

Question 2:

Solve the following equation using the Quadratic Formula: $2x^2 + 7x + 5 = 0$.

Solution:

Comparing $2x^2 + 7x + 5 = 0$ with $ax^2 + bx + c = 0$, we get:

$a = 2, b = 7, c = 5$

Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(5)}}{2(2)}$

$x = \frac{-7 \pm \sqrt{49 - 40}}{4}$

$x = \frac{-7 \pm \sqrt{9}}{4} = \frac{-7 \pm 3}{4}$

Case 1: $x = \frac{-7 + 3}{4} = \frac{-4}{4} = -1$

Case 2: $x = \frac{-7 - 3}{4} = \frac{-10}{4} = -2.5$

Roots: $x = -1, -2.5$

Question 3:

Determine the nature of the roots of the quadratic equation: $3x^2 - 4x + 1 = 0$.

Solution:

Here, $a = 3, b = -4, c = 1$

Discriminant ($D$) $= b^2 - 4ac$

$D = (-4)^2 - 4(3)(1)$

$D = 16 - 12 = 4$

Since $D > 0$ and $D$ is a perfect square, the roots are real, rational, and unequal.

Question 4:

Solve $x^2 + 6x + 9 = 0$.

Solution:

Given: $x^2 + 6x + 9 = 0$

This is in the form of $(a + b)^2 = a^2 + 2ab + b^2$.

$(x)^2 + 2(x)(3) + (3)^2 = 0$

$(x + 3)^2 = 0$

$x + 3 = 0$

$x = -3, -3$ (Equal roots)

Question 5:

Find the value of $k$ if one root of the quadratic equation $kx^2 - 14x + 8 = 0$ is $2$.

Solution:

Since $x = 2$ is a root, it must satisfy the equation.

$k(2)^2 - 14(2) + 8 = 0$

$4k - 28 + 8 = 0$

$4k - 20 = 0$

$4k = 20$

$k = 5$

Question 6:

The sum of two numbers is 15 and the sum of their reciprocals is $3/10$. Find the numbers.

Solution:

Let the numbers be $x$ and $15 - x$.

According to the condition: $\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10}$

$\frac{15 - x + x}{x(15 - x)} = \frac{3}{10}$

$\frac{15}{15x - x^2} = \frac{3}{10}$

$150 = 3(15x - x^2)$

$50 = 15x - x^2$ (dividing by 3)

$x^2 - 15x + 50 = 0$

$(x - 10)(x - 5) = 0$

The numbers are 10 and 5.

Question 7:

Solve: $x^2 - 2x - 15 = 0$

Solution:

$x^2 - 5x + 3x - 15 = 0$

$x(x - 5) + 3(x - 5) = 0$

$(x - 5)(x + 3) = 0$

$x = 5$ or $x = -3$

Question 8:

Form a quadratic equation whose roots are $4$ and $-3$.

Solution:

Sum of roots ($\alpha + \beta$) $= 4 + (-3) = 1$

Product of roots ($\alpha\beta$) $= 4 \times (-3) = -12$

Equation: $x^2 - (\text{Sum})x + (\text{Product}) = 0$

$x^2 - (1)x + (-12) = 0$

$x^2 - x - 12 = 0$

Question 9:

Solve $4x^2 - 20x + 25 = 0$ using factorization.

Solution:

$4x^2 - 10x - 10x + 25 = 0$

$2x(2x - 5) - 5(2x - 5) = 0$

$(2x - 5)(2x - 5) = 0$

$2x - 5 = 0 \Rightarrow 2x = 5$

$x = 5/2$

Question 10:

Solve: $x + \frac{1}{x} = 2.5$

Solution:

Multiply the whole equation by $x$:

$x^2 + 1 = 2.5x$

$x^2 - 2.5x + 1 = 0$

Multiply by 2 to remove decimals: $2x^2 - 5x + 2 = 0$

$2x^2 - 4x - x + 2 = 0$

$2x(x - 2) - 1(x - 2) = 0$

$(x - 2)(2x - 1) = 0$

$x = 2$ or $x = 1/2$