Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 09th Sep 2024
Total views: 410.4k
Views today: 4.10k
Verified
410.4k+ views
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.

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.