Class 10- Find the roots of the quadratic equation

Quadratic equations are an important part of mathematics and are used by in many branches of science and engineering. There are many methods which one can follow to find roots of a quadratic equation.

One such method is finding roots of quadratic equation by completing the square.

Since we have already studied the completing the square method, we will now learn about quadratic equation by factorization. Understanding the concept of finding the roots of a quadratic equation  is simple. Here we go.

We already know that the standard form of quadratic equation is ax2 + bx + c = 0, where a ≠ 0 and a, b, and c are constants. Considering the same equation, if we replace ‘x’ by 1, on the LHS of the equation, we will get (a X 1)2 + (b X 1) + c = 0. So, it means that 1 is a root of the equation  ax2 + bx + c = 0.

In other words, the real number ‘a’ is called as the root of the quadratic equation, where x is replaced by ‘a’. Therefore, if we compute ‘a’ with the given equation, we arrive at a2 + ba + c = 0, which is a solution of the quadratic equation.

Method to find the roots of quadratic equation

  1. Let the quadratic equation be ax2 + bx + c = 0.
  2. First we need to factorize the expression ax2 + bx + c.
  3. After factorizing, the expression will be in the form (x+p)(x+q).
  4. Equate each part to zero and get the roots.

Let us see some examples on finding the roots of quadratic equations.

Examples on finding the roots of the quadratic equation

1) Find the roots of the following quadratic equation   x2 – 3x – 10 = 0.

Solution:  Split the middle term -3x as -5x + 2x because (-5x)(2x) = -10x2 = (-x2)10

x2 – 5x + 2x – 10 = 0

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

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

Equating each part to zero, we get

x = -2, 5.

2) Find two numbers whose sum is 27 and product is 182.

Solution:                  Let x and y be the two numbers.

According to the problem,

x+y = 27 …..(1)

xy = 182 …..(2)

From equation (1), we have x+y = 27

y = 27 – x

Substitute this in equation (2)

x (27-x) = 182

27x – x2 = 182

x2 -27x + 182 = 0

Split middle term -27x as -14x -13x because (-14x)(-13x) = 182x2 = (182)(x2)

x2 – 14x -13x + 182 = 0

x(x-14) – 13(x – 14) = 0

(x-13)(x-14) = 0

Equating each part to zero, we get

x = 13, 14.

Therefore y = 14, 13

Important Note:  we can check the roots correct or not by substituting the roots in the quadratic equation.

3) Find two consecutive positive integers, sum of whose squares is 365.

Solution:                   Let x and x+1 be the consecutive positive integers.

Sum of squares of these integers is 365

x2 + (x+1)2 = 365

x2 + x2 + 2x + 1 = 365

2x2 + 2x + 1 – 365 = 0

2x2 + 2x – 364 = 0

x2 + x – 182 = 0

Split middle term x as 14x -13x because (14x)(-13x) = -182x2 = (-182)(x2)

x2 +14x -13x – 182 = 0

x(x+14) – 13(x+14) = 0

(x-13)(x+14) = 0

Equating each part to zero, we get

x = -14,13

Here x ≠ -14, given that the numbers are positive integers

Therefore x = 13.

The consecutive number is 14.

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