Force

Quadratic Equation

Unit: 7

Book Icon Class 10: Mathematics

Quadratic Equation, Methods of solving quadratic equations, Worked out examples, SEE Questions with Solutions

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in the form: \(ax^2 + bx + c = 0\)

where:

- \( a, b, c \) are constants, with \( a \neq 0 \), \( x \) represents the variable.

It has only one variable with degree 2. The quadratic equation has a single variable.

 

Solution of Quadratic Equation

The solution of a quadratic equation \(ax^2 + bx + c = 0\) refers to the values of \(x\) that satisfy the equation. These values are also called roots or zeroes of the equation. It has two roots, meaning that the quadratic equation has two values of \(x\) that satisfy the equation. The quadratic equation has two roots.

Example:

Q. Find the roots of the quadratic equation \(x^2 + 5x + 6 = 0\).

\(\begin{align} & \text{Solution: } \\ & x^2 + 5x + 6 = 0 \\ & or, x^2 + (3 + 2)x + 6 = 0 \\ & or, x^2 + 3x + 2x + 6 = 0 \\ & or, x(x + 3) + 2(x + 3) = 0 \\ & \therefore (x + 3) (x + 2) = 0 \end{align}\)

Hence, the roots of the equation \(x^2 + 5x + 6 = 0\) are \(x = -3\) and \(x = -2\). If you put these values in the original equation \(x^2 + 5x + 6 = 0\) the result will be \(0 = 0\). Hence, the solution satisfies the equation.

 

Methods to Solve a Quadratic Equation

There are several methods to solve quadratic equations. In this class, we will learn methods like factoring, completing the square, and using the quadratic formula. Each of them is discussed below with an example question.

1. Factoring 

   - Express the quadratic equation in factored form: 

     \((x - p)(x - q) = 0\)

   - Solve for \( x \) by setting each factor to zero: 

     \(x - p = 0 \quad \text{or} \quad x - q = 0\)

 

2. Completing the Square 

   - Rewrite the equation in the form: 

     \((x - h)^2 = k\)

   - Solve for \( x \) by taking the square root.

 

3. Quadratic Formula 

   - Compare the given equation with the standard form \(ax^2 + bx + c = 0\) and get the values of the constants a, b and c. Then, use the following formula: 

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

   - Note: This method works for all quadratic equations, even if they are not factorable.

 

Example Question with Solution

Q. Solve the equation \(x^2 – 6x + 9 = 0\) by

a. Factorization Method. [2U]

b. Completing square method. [2U]

c. Using Formula Method. [2U]

 

\(\begin{aligned} & \text{Solution: } \\ & \text{a. Factorization Method} \\ & or, x^2 – 6x + 9 = 0 \\ & or, x^2 – (3 + 3)x + 9 = 0 \\ & or, x^2 – 3x - 3x + 9 = 0 \\ & or, x(x – 3) – 3(x – 3) = 0 \\ & or, (x – 3) (x-3) = 0 \\ & \text{Either, } (x - 3) = 0 \quad \text{OR, } (x - 3) = 0 \\ & \therefore \boxed{x = 3} \quad \text{OR, } \therefore \boxed{x = 3} \\ & \text{Hence, the solutions are, x = 3 and 3. } \end{aligned}\)

 

\(\begin{aligned} & \text{Solution: } \\ & \text{b. Completing Square Method } \\ & or, x^2 – 6x + 9 = 0 \\ & or, x^2 – 2 . x. 3 + 3^2 = 0 \\ & or, (x - 3)^2 = 0 \\ & or, (x – 3) (x-3) = 0 \\ & \text{Either, } (x - 3) = 0 \quad \text{OR, } (x - 3) = 0 \\ & \therefore \boxed{x = 3} \quad \text{OR, } \therefore \boxed{x = 3} \\ & \text{Hence, the solutions are, x = 3 and 3. } \end{aligned}\)

 

\(\begin{aligned} & \text{Solution: } \\ & \text{c. Using Formula Method } \\ & or, x^2 – 6x + 9 = 0 \\ & Now, \text{Comparing it with } ax^2 + bx + c = 0 \\ & \text{Then, } a = 1, b = -6 \text{ and } c = 9 \\ & \text{We have, } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ & or, x =  \frac{6 \pm \sqrt{(-6)^2 – 4 \times 1 \times 9}}{2 \times 1} \\ & or, x = \frac{6 \pm \sqrt{36 - 36}}{2} \\ & or, x = \frac{6 \pm 0}{2} \\ & or, x = \frac{6}{2} \\ & \therefore \boxed{x = 3} \\ & \text{Hence, the solutions are, x = 3 and 3. } \end{aligned}\)

 

Practice Question Solving Quadratic Equation

Q1. Solve the equation \(2x^2 + x - 6 = 0\) by

a. Factorization Method. [2U]

b. Completing square method. [2U]

c. Using Formula Method. [2U]

 

Word Problems Related to Quadratic Equation

In this section, we will focus on solving real-world problems expressed in words. Each problem will be carefully analyzed to understand the given information and determine the appropriate mathematical equations. Once the equations are formed, they will be solved using any of the methods discussed in the previous section.

 

🔹 Steps for Solving Word Problems

1. Read the problem carefully — Understand what the question is asking and identify all the important information.

2. Define variables — Let symbols (like x, y) represent the unknown quantities.

3. Translate the words into equations — Use the given relationships or conditions to form mathematical equations.

4. Solve the equations — Apply an appropriate method (factorization, completing the square, or quadratic formula).

5. Interpret the results — Substitute the values back into the context of the problem.

6. Check your answer — Verify that the solution makes sense and satisfies the conditions of the problem.

 

Example Question

Q. The numerical product of the present ages of two sisters Rita and Gita is 160. Four years ago, Rita was twice as old as Gita. [SEE 2080 GI]

a. If the age of Gita 4 years ago was x years, then what was the age of Rita? Write it. [1K]

b. Find the present ages of Rita and Gita. [3A]

c. If both live on, then what will be their ages after 10 years? Find it. [1A]

Solution:

(a) If the age of Gita 4 years ago was x years, then what was the age of Rita? Write it.
\(\rightarrow \) Let the age of Gita 4 years ago = \( x \) years.
Then, the age of Rita 4 years ago = \( 2x \) years.

(b) Find the present ages of Rita and Gita.

\(\begin{aligned} & \rightarrow \text{Gita’s present age} = x + 4, \\ & \text{Rita’s present age} = 2x + 4. \\ & \text{Now, According to the question,} \\ & or, (2x + 4)(x + 4) = 160 \\ & or, 2x^2 + 8x + 4x + 16 = 160 \\ & or, 2x^2 + 12x + 16 - 160 = 0 \\ & or, 2x^2 + 12x - 144 = 0 \\ & or, x^2 + 6x - 72 = 0 \\ & or, (x + 12)(x - 6) = 0 \\ & \therefore \boxed{x = -12} \text{ or } \therefore \boxed{x = 6} \\ & \text{Since age cannot be negative, } x = 6. \\ & \text{Hence,} \\ & \text{Gita’s present age} = x + 4 = 6 + 4 = 10 \text{ years,} \\ & \text{Rita’s present age} = 2x + 4 = 12 + 4 = 16 \text{ years.} \end{aligned}\)

(c) If both live on, then what will be their ages after 10 years? Find it.

\(\begin{aligned} & \rightarrow \text{After 10 years:} \\ & \text{Gita’s age} = 10 + 10 = 20 \text{ years,} \\ & \text{Rita’s age} = 16 + 10 = 26 \text{ years.} \end{aligned}\)

Final Answers:

  • (a) Rita’s age 4 years ago = \( 2x \) years
  • (b) Gita’s present age = \( 10 \) years, Rita’s present age = \( 16 \) years
  • (c) After 10 years: Gita = \( 20 \) years, Rita = \( 26 \) years

 

Practice Question Word Related Quadratic Equation

Q1. The product of two consecutive positive numbers is 420. [SEE 2080 KoP]

a. What are the numbers? Find it. [2U]

b. What should be subtracted from the numbers to get their product as 182? Find it. [2A]

 

Share Now

Share to help more learners!

Feedback     Report an Issue

© Hupen. All rights reserved.

Handcrafted with in Nepal by Hupen Design