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Question: For showing the quadratic equation by ‘completing square method’. Find the third term : \[{x^2} + 8x...

For showing the quadratic equation by ‘completing square method’. Find the third term : x2+8x=15{x^2} + 8x = - 15

Explanation

Solution

Method of completing the square is a method which is used to find out the roots of a quadratic equation Having Degree 2. Following are the steps to solve a quadratic equation by the method of completing the square. Do add and subtract the square of half of x, then separate the variables, constant, and solve them.

Complete step-by-step answer:
We are given the quadratic equation x2+8x=15{x^2} + 8x = - 15
\Rightarrow $$$${x^2} + 8x + 15 = 0...........(1)
For this matter add and subtract the term square of half the coefficient of xx here coefficient of is 88, half of 88 is for and the square of 44 is 1616.
Therefore add and subtract 1616 in (1)(1).
\Rightarrow $$$${x^2} + 8x + 16 + 15-16 = 0
Here combining the first three terms and using formula a2+b2+2ab=(a+b)2{a^2} + {b^2} + 2ab = {(a + b)^2}.
\Rightarrow $$$${(x)^2} + (2x)(4x) + {(4)^2} - 1 = 0
\Rightarrow $$$${(x + 4)^2} - 1 = 0
\Rightarrow $$$${(x + 4)^2} = 1
Taking square root on both sides x+4=±1x + 4 = \pm 1
which given 2 values x+4=1x + 4 = 1 and x+4=1x + 4 = - 1
\Rightarrow $$$$x = 1 - 4 and x=14x = - 1 - 4
So the value of x=3x = - 3 and x=5x = - 5
Which given two values 3,5 - 3, - 5
Therefore we get two values of variable xx as 3 - 3 and 5 - 5

Therefore on solving the quadratic equation by the method of completing squares, we should get two values of variables and those two values both variables are 3 - 3 and 5. - 5.

Note: There are three methods to solve quadratic equations. A quadratic equation is of the type ax2+bx+c=0a{x^2} + bx + c = 0 where a, b, c,a,{\text{ }}b,{\text{ }}c, are constants (whose value is fixed) and N is the variable(whose value varies) and the three methods of solving quadratic equations are

Middle term splitting
Method of completing squares
Method of discriminant
And with any of these three methods, the two values of the variable in each metal is the same as when solving with the other two methods.