Question

How do you solve using the square root method ${x^2} + 5 = 86$?

Answer 1

Hint: In order to solve the given quadratic equation, first simplify the equation by subtracting value$5$ from both sides of the equation. Now transpose the square from LHS to RHS it will become square root. Remember to include $ \pm $in the RHS side.

Complete step by step answer:
We are given a quadratic equation in variable $x$ as
${x^2} + 5 = 86$
Simplifying the equation by subtracting $5$ from both the sides of the equation to eliminate 5 from the left-hand side, we have
 $
  {x^2} + 5 - 5 = 86 - 5 \\
  {x^2} = 81 \\
 $
Now transposing square form left-hand side of the equation to right-hand side , it will become square root
$x = \pm \sqrt {81} $
As we know $81$ is a perfect square of $9$as$9 \times 9 = 81$, so $\sqrt {81} = 9$
$x = \pm 9$
$x = 9, - 9$
Therefore, the solution of the given quadratic equation ${x^2} + 5 = 86$ is equal to $x = 9\,and\,x = - 9$.
Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$

Hence the factors will be $(x - x1)\,and \,(x - x2)\,$.

Additional Information:
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

Note: 1. One must be careful while calculating the answer as calculation error may occur
2.$ \pm $ is included while taking square root as every positive number has two square roots, one is positive and other is negative.
3.Since the equation is quadratic, we obtained two solutions for variable $x$
4.Degree of equation is equal to the number of solutions of that equation.