Quadratic Equation Questions and Their Solutions – Mindspark

Quadratic Equation Questions

Quadratic equations are an essential part of algebra, and we must be familiar with them and the methods of solving quadratic equation questions. 

Before solving quadratic equation questions, let us have a quick recap of the quadratic equation concept and formulas.

Definition of Quadratic Equations

A quadratic equation is represented in the form of (ax² + bx + c) where x is the variable, and a, b and c are real numbers or constants. It is a polynomial in which the highest power of the variable is 2. 

In the quadratic equation’s standard form, the value of a can not be zero. If ‘a’ is zero, it will remove the ‘x²’ term, and the equation will not remain quadratic.

Some examples of quadratic equations are:

45x² + 3x + 4, where a = 45, b = 3 and c = 4 (when we compare it with ax² + bx + c).

Similarly, in the quadratic equation (- 4x² + 54x – 40), we have a = – 4, b = 54 and c = – 40.

Roots of a Quadratic Equation

The roots of the quadratic equation are the values of x for which ax² + bx + c = 0. Since the degree of a quadratic equation is 2, we obtain two roots.

Quadratic Equation Formula

The quadratic equation formula is also known as the Sridharacharya Formula. It is a method to find the roots of the equation. This formula is given below:

(b² – 4ac) is called the discriminant of the equation and is represented by D.

Therefore D = (b² – 4ac)

We can define the nature of roots based on the discriminant value. There are 3 possible conditions given below:

• two distinct real roots, if D > 0.
• two real roots and both equal, if D = 0.
• no real roots, if D < 0.

Quadratic Equations Questions and Solutions

Now let us see some quadratic equation problems and their solutions.

1. Check if x(x + 2) + 6 = (x + 2) (x – 2) is in the form of quadratic equation.

Solution: Given,

x(x + 2) + 6 = (x + 2) (x – 2)

⇒ x² + 2x + 6 = x² – 2²    [By algebraic identities]

On Simplifying the equation

2x + 6 = – 4

⇒ 2x + 10 = 0

Since this equation is not in the form of ax² + bx + c, it is not a quadratic equation.

2. Find the roots of the equation 2x² – 8x + 8 = 0 using factorisation.

Solution:  Given,

2x² – 8x + 8 = 0

⇒ 2x² – 4x – 4x + 8 = 0

⇒ 2x(x – 2) – 4(x – 2) = 0

⇒ (2x – 4) (x – 2) = 0

So, (2x – 4) = 0, and (x – 2) = 0

If 2x – 4 = 0; x = 2,

and if (x – 2) = 0, we get x = 2

Therefore, there are two equal roots of the given equation, and the root is 2.

3. Solve the quadratic equation x² + 4x – 5 = 0.

Solution:

x² + 4x – 5 = 0

⇒ x² – x + 5x – 5 = 0

⇒ x(x – 1) + 5(x – 1) = 0

⇒ (x – 1)(x + 5) = 0

So, (x – 1) = 0 and (x + 5) = 0.

If x – 1 = 0; x = 1.

And if x + 5 = 0; x = -5

Therefore, the roots of the given quadratic equation are -5 and 1.

4. Solve the quadratic equation 2x² + x – 528 = 0, using quadratic formula.

Solution: Comparing the equation with standard equation ax² + bx + c = 0.

We find, a = 2, b = 1, and c = -528.

By using the quadratic formula:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

x=\frac{-1 \pm \sqrt{1^{2}-(4\times{(2)}\times{(-528))}}}{2 \times 2}

⇒ x=\frac{-1 \pm \sqrt{1-(4\times(2)\times (-528))}}{4}

⇒ x=\frac{-1 \pm \sqrt{1+4224}}{4}

⇒ x=\frac{-1 \pm \sqrt{4225}}{4}

⇒ x=\frac{-1 \pm 65}{4}

So, x=\frac{-1 + 65}{4}\text{ and }x=\frac{-1-65}{4}

⇒ x=\frac{64}{4}\text{ and }x=\frac{-66}{4}

⇒ x=16\text{ and }x=\frac{-33}{2}

Therefore, the roots of the equation are 16 and -33/2.

5. What is the discriminant of the equation 3x² – 2x +1/3  = 0.

Solution: Comparing the equation with standard equation ax² + bx + c = 0, 

a = 3, b = – 2, and c = ⅓.

Formula for discriminant, D = b² – 4ac

D = (-2)² – 4×(3)×(⅓) 

⇒ D = 4 – 4

⇒ D = 0

Examples

1. Find the roots of the quadratic equation x^{2}-45 x+324=0

Solving this quadratic equation- 

⇒ x(x-36)-9(x-36)=0

⇒ (x-9)(x-36)=0

So, x = 9 and 36.

2. Solve the quadratic equation x² + 5x + 7 = 21.

The given polynomial or quadratic equation is

x² + 5x + 7 = 21

Using factorization method, 

x² + 5x + 7 – 21 = 0

⇒ x² + 5x – 14 = 0

⇒ x² – 2x + 7x – 14 = 0

⇒ x(x – 2) + 7(x – 2) = 0

⇒ (x – 2)(x + 7) = 0

So, (x – 2) = 0, and (x + 7) = 0 

The roots (solution) of the quadratic equation, x = 2, -7.