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Question: How do you solve using the square root method \({x^2} + 5 = 86\)?...

How do you solve using the square root method x2+5=86{x^2} + 5 = 86?

Explanation

Solution

In order to solve the given quadratic equation, first simplify the equation by subtracting value55 from both sides of the equation. Now transpose the square from LHS to RHS it will become square root. Remember to include ±\pmin the RHS side.

Complete step by step answer:
We are given a quadratic equation in variable xx as
x2+5=86{x^2} + 5 = 86
Simplifying the equation by subtracting 55 from both the sides of the equation to eliminate 5 from the left-hand side, we have
x2+5−5=86−5 x2=81  {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=±81x = \pm \sqrt {81}
As we know 8181 is a perfect square of 99as9×9=819 \times 9 = 81, so 81=9\sqrt {81} = 9
x=±9x = \pm 9
x=9,−9x = 9, - 9
Therefore, the solution of the given quadratic equation x2+5=86{x^2} + 5 = 86 is equal to x=9 and x=−9x = 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=−b+b2−4ac2ax1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} and x2=−b−b2−4ac2ax2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}
x1,x2 are root to quadratic equation ax2+bx+ca{x^2} + bx + c

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

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of ax2+bx+ca{x^2} + bx + cwhere xxis the unknown variable and a,b,c are the numbers known where a≠0a \ne 0.If a=0a = 0then 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=b2−4acD = {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 xx
4.Degree of equation is equal to the number of solutions of that equation.