Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you solve by completing the square ${x^2} + 18x + 80 = 0$?

seo-qna
Last updated date: 20th Jun 2024
Total views: 375.3k
Views today: 9.75k
Answer
VerifiedVerified
375.3k+ views
Hint: To determine the solution to the above equation using completing the square method, first divide the equation by the coefficient of ${x^2}$ to convert into ${x^2} + bx$ form and add ${\left( {\dfrac{b}{2}} \right)^2}$
${\left( {\dfrac{b}{2}} \right)^2}$to make ${x^2} + bx + {\left( {\dfrac{b}{2}} \right)^2}$ which is a perfect square of ${\left( {x + \dfrac{b}{2}} \right)^2}$. Now apply the formula $({A^2} - {B^2}) = (A + B)(A - B)$ to factorise it further and find the required values of $x$.

Complete step-by-step solution:
We are given a quadratic equation ${x^2} + 18x + 80 = 0$
First divide the whole equation with the coefficient of ${x^2}$ to obtain the form${x^2} + bx$in the equation only if the coefficient is not equal to 1.
Since in our case the coefficient is already equal to one, so no need to divide.
${x^2} + 18x + 80 = 0$
We are going to rearrange ${x^2} + bx + e$in order to complete it into a square by adding${\left( {\dfrac{b}{2}} \right)^2}$to make ${x^2} + bx + {\left( {\dfrac{b}{2}} \right)^2}$which is a perfect square of ${\left( {x + \dfrac{b}{2}} \right)^2}$where b is the coefficient of x .
\[{\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{{18}}{2}} \right)^2} = {\left( 9 \right)^2} = 81\]
Therefore we have to add $81$ in the equation to make it a square but we can’t simply add $81$ on LHS as this will disturb the balance, so we will add $81$on both sides of the equation. Our equation will now become .
\[ \Rightarrow {x^2} + 18x + 81 + 80 = 81\]
\[{x^2} + 18x + 81\] can be written as ${\left( {x + 9} \right)^2}$
$
   \Rightarrow {\left( {x + 9} \right)^2} + 80 = 81 \\
   \Rightarrow {\left( {x + 1} \right)^2} - 81 + 80 = 0 \\
   \Rightarrow {\left( {x + 1} \right)^2} - 1 = 0 \\
 $
$1$can be written as \[{1^2}\]
$ \Rightarrow {\left( {x + 9} \right)^2} - {1^2} = 0$
Using the identity $({A^2} - {B^2}) = (A + B)(A - B)$ Where A is$x + 9$ and B as $1$
$
   \Rightarrow {\left( {x + 9} \right)^2} - {1^2} = 0 \\
   \Rightarrow \left( {x + 9 + 1} \right)\left( {x + 9 - 1} \right) = 0 \\
   \Rightarrow \left( {x + 10} \right)\left( {x + 8} \right) = 0 \\
  x + 10 = 0 \\
   \Rightarrow x = - 10 \\
  x + 8 = 0 \\
   \Rightarrow x = - 8 \\
 $
Therefore, the solution to ${x^2} + 18x + 80 = 0$ is equal to $x = - 8, - 10$

Note:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
-If D is equal to zero, then both of the roots will be the same and real.
-If D is a positive number then, both of the roots are real solutions.
-If D is a negative number, then the root are the pair of complex solutions