Quadratic equation questions are provided here for Class 10 students. A quadratic equation is a second-degree polynomial which is represented as ax^{2} + bx + c = 0, where a is not equal to 0. Here, a, b and c are constants, also called coefficients and x is an unknown variable. Also, learn Quadratic Formula here.

Solving the problems based on quadratics will help students to understand the concept very well and also to score good marks in this section. All the questions are solved here step by step with detailed explanation. In this article, we will give the definition and important formula for solving problems based on quadratic equations. The questions given here is in reference with CBSE syllabus and NCERT curriculum. Also, learn quadratic equations for class 10 here.

**Definition of Quadratic Equation**

Usually, the quadratic equation is represented in the form of ax^{2}+bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0. Here, a and b are the coefficients of x^{2} and x, respectively. So, basically a quadratic equation is a polynomial whose highest degree is 2. Let us see some examples:

- 3x
^{2}+x+1, where a=3, b=1, c=1 - 9x
^{2}-11x+5, where a=9, b=-11, c=5

**Roots of Quadratic Equations:**

If we solve any quadratic equation, then the value we obtained are called the roots of the equation. Since the degree of the quadratic equation is two, therefore we get here two solutions and hence two roots.

There are different methods to find the roots of quadratic equation, such as:

- Factorisation
- Completing the square
- Using quadratic formula

**Quadratic Equation Formula:**

The quadratic formula to find the roots of the quadratic equation is given by:

**\(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)**

Where, b^{2}-4ac is called discriminant of the equation.

Based on the discriminant value, there are three possible conditions, which defines the nature of roots as follows:

- two distinct real roots, if b
^{2}– 4ac > 0 - two equal real roots, if b
^{2 }– 4ac = 0 - no real roots, if b
^{2}– 4ac < 0

## Quadratic Equations Problems and Solutions

**1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with. **

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x^{2} – 200 + 5x = 124

– x^{2} + 45x – 200 = 124

x^{2} – 45x + 324 = 0

This represents the problem mathematically.

**2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.**

Solution: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

x^{2}+x+8 = x^{2}-2^{2} [By algebraic identities]

Cancel x^{2} both the sides.

x+8=-4

x+12=0

Since, this expression is not in the form of ax^{2}+bx+c, hence it is not a quadratic equation.

**3. Find the roots of the equation 2x ^{2} – 5x + 3 = 0 using factorisation.**

Solution: Given,

2x^{2} – 5x + 3 = 0

2x^{2} – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

So,

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

**4. Solve the quadratic equation 2x ^{2}+ x – 300 = 0 using factorisation.**

Solution: 2x^{2}+ x – 300 = 0

2x^{2} – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

So,

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of given equation.

**5. Solve the equation x ^{2}+4x-5=0, using completing the square method.**

Solution:

x^{2} + 4x – 5 = 0

x^{2}+8/2x-5 = 0

x^{2}+4/2x+4/2x-5 = 0

x^{2}+2x+2x-5 = 0

(x + 2) x + 2 × x – 5= 0

(x + 2) x + 2 × x + 2 × 2 – 2 × 2 – 5= 0

(x + 2) x + (x + 2) × 2 – 2 × 2 – 5 = 0

(x+2) (x+2) -2^{2} – 5 = 0

(x+2)^{2} – 2^{2} – 5 = 0

(x+2)^{2} – 4 – 5 = 0

(x+2)^{2} – 9 = 0

(x+2)^{2} – 3^{2} = 0

(x+2+3) (x+2-3) = 0 [By algebraic identities]

(x+5) (x-1) = 0

Therefore,

x=-5 & x=1

**6. Solve the quadratic equation 2x ^{2} + x – 528 = 0, using quadratic formula.**

Solution: If we compare it with standard equation, ax^{2}+bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

\(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)Now putting the values of a,b and c.

\(x=\frac{-1 \pm \sqrt{1+4(2)(528)}}{4}=\frac{-1 \pm \sqrt{4225}}{4}=\frac{-1 \pm 65}{4}\)x=64/4 or x=-66/4

x=16 or x=-33/2

**7. Find the roots of x ^{2} + 4x + 5 = 0, if any exist, using quadratic formula.**

Solution: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b^{2}-4ac = 4^{2}-4.1.5 = 16-20 = -4

Snce square root of -4 will not give real number. Hence there is no real roots for the given equation.

**8. Find the discriminant of the equation: 3x ^{2}-2x+⅓ = 0.**

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b^{2} – 4ac

D = (-2)^{2}-4.3.(⅓)

D = 4-4

D=0

## Video Lesson

### Quadratic Equation Worksheet

### Practice Questions

Solve these quadratic equations and find the roots.

- x
^{2}-5x-14=0 [Answer: x=-2 & x=7] - X
^{2}= 11x -28 [Answer: x=4 & x = 7] - 6x
^{2}– x = 5 [Answer: x=-⅚ & x = 1] - 12x
^{2}= 25x [Answer: x=0 & x=25/12]

## Frequently Asked Questions – FAQs

### What is a quadratic equation?

^{2}+bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0.

### What are the examples of quadratic equations?

5x

^{2}– x + 6 = 0

x

^{2}+ 8x + 2 = 0

-x

^{2}+ 6x + 18 = 0

x

^{2}– 4 = 0

### What is the formula for quadratics?

x = [-b ± √(b

^{2}-4ac)]/2a

### What are the methods to solve the quadratic equation?

Factorisation

Square root property

Completing the square

Using the quadratic formula

### What are the roots of quadratic equation?

x1 = [-b + √(b

^{2}-4ac)]/2a and

x2 = [-b – √(b

^{2}-4ac)]/2a

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