Question

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

Answer 1

Hint: 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 \[{x^2} + 8x = - 15\]
\[ \Rightarrow \]\[{x^2} + 8x + 15 = 0...........(1)\]
For this matter add and subtract the term square of half the coefficient of \[x\] here coefficient of is \[8\], half of \[8\] is for and the square of \[4\] is \[16\].
Therefore add and subtract \[16\] in \[(1)\].
\[ \Rightarrow \]\[{x^2} + 8x + 16 + 15-16 = 0\]
Here combining the first three terms and using formula \[{a^2} + {b^2} + 2ab = {(a + b)^2}\].
\[ \Rightarrow \]\[{(x)^2} + (2x)(4x) + {(4)^2} - 1 = 0\]
\[ \Rightarrow \]\[{(x + 4)^2} - 1 = 0\]
\[ \Rightarrow \]\[{(x + 4)^2} = 1\]
Taking square root on both sides \[x + 4 = \pm 1\]
which given 2 values \[x + 4 = 1\] and \[x + 4 = - 1\]
\[ \Rightarrow \]\[x = 1 - 4\] and \[x = - 1 - 4\]
So the value of \[x = - 3\] and \[x = - 5\]
Which given two values \[ - 3, - 5\]
Therefore we get two values of variable \[x\] as \[ - 3\] and \[ - 5\]

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 \[ - 3\] and \[ - 5.\]

Note: There are three methods to solve quadratic equations. A quadratic equation is of the type \[a{x^2} + bx + c = 0\] where \[a,{\text{ }}b,{\text{ }}c,\] 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.