Question

Find the Roots of 5x2_6x-2-oby the
method of competing the square.​

Answer 1

Method:

Let the equation be = ax^2 + bx + c = 0

• Make the coefficient of x^2 as equal to 1. This can be done by dividing the whole equation by the a, applied that a >< 1.

The equation now becomes: x^2 + (b/a)x + (c/a) = 0

• Divide (b/a) by 2. The coefficient of x now becomes = b/2a. Now add and subtract (b/2a)^2 in the equation.

The equation becomes: x^2 + (b/a)x + (b/2a)^2 - (b/2a)^2 + (c/a) = 0

• See the first 3 terms. They make up to the equation: a^2 + 2ab + b^2 = (a + b)^2

Equation becomes: (x + b/2a)^2 + (b^2/4a^2) + (c/a) = 0

• Now solve the equation. You will get the values.

Solution:

Equation: 5x^2 - 6x - 2 = 0

Step-wise:

• Make the coefficient of x^2 as equal to 1. This can be done by dividing the whole equation by the a, applied that a >< 1.

The equation now becomes: 5x^2 - (6/5)x - (2/5) = 0

• Divide (b/a) by 2. The coefficient of x now becomes = b/2a. Now add and subtract (b/2a)^2 in the equation.

The equation becomes: 5x^2 - (6/5)x + (6/10)^2 - (6/10)^2 + (2/5) = 0

Simplify it: 5x^2 - (6/5)x + (3/5)^2 - (3/5)^2 + (2/5) = 0

• See the first 3 terms. They make up to the equation: a^2 + 2ab + b^2 = (a + b)^2

Equation becomes: (x + 3/5)^2 - (9/25) - (2/5) = 0

• Now solve the equation. You will get the values.

Equation: (x + 3/5)^2 = 19/25

= x + 3/5 = (+_/19/5) , (-_/19/5)

= x = (3 + _/19)/5 , (3 - _/19)/5

Note: "_/" represents square root sign.