Question 241004
In general, there are three ways to solve quadratic equations:<ul><li>Complete the square and find the square root fo each side.</li><li>Get one side equal to zero and<ul><li>Factor; or</li><li>Use the Quadratic Formula</li></ul></li></ul>Most the the time completing the square is the most difficult. But when there is no x term, just {{{x^2}}} terms, then completing the square is fairly easy. Since all of your problems have no x terms, we will complete the square.<br>
To complete the square, get the x terms on one side and form a perfect square. Then find the square root of each side and solve.<br>
1) {{{x^2=60}}}
The x terms are already on the left and {{{x^2}}} is already a perfect square. So we will find the square root of each side:
{{{sqrt(x^2) = sqrt(60)}}}
A very common mistake is to think that {{{sqrt(x^2) = x}}}. It is not. {{{sqrt(x^2) = abs(x)}}}! So our equation simplifies to:
{{{abs(x) = sqrt(60)}}}
Solving an absolute value equation requires two equations:
{{{x = sqrt(60)}}} or x = {{{-sqrt(60)}}}
Since {{{sqrt(60) = sqrt(4*15) = sqrt(4)*sqrt(15) = 2sqrt(15)}}} our final answer is
{{{x = 2sqrt(15)}}} or {{{x = -2sqrt(15)}}}<br>
On the remainder of the problems I will be using the same steps:<ol><li>Complete the square<ol><li>Gather the x terms on one side</li><li>Form a perfect square</li></ol></li><li>Find the square root of each side</li><li>Solve</li></ol>So there will be fewer comments on the rest of the problems.<br>

2) {{{x^2-9=16}}}
Add 9 to both sides
{{{x^2 = 25}}}
{{{sqrt(x^2) =  sqrt(25)}}}
{{{abs(x) = 5}}}
{{{x = 5}}} or {{{x = -5}}}<br>

3) {{{4x^2+7=23}}}
Subtract 7 from each side
{{{4x^2 = 16}}}
Divide both sides by 4
{{{x^2 = 4}}}
{{{sqrt(x^2) = sqrt(4)}}}
{{{abs(x) = 2}}}
{{{x - 2}}} or {{{x = -2}}}<br>

4) {{{x^2/6-2=0}}}
Add 2 to both sides
{{{x^2/6 = 2}}}
Multiply both sides by 6
{{{x^2 = 12}}}
{{{sqrt(x^2) = sqrt(12)}}}
{{{abs(x) = sqrt(12)}}}
{{{x = sqrt(12)}}} or {{{x = -sqrt(12)}}}
Since {{{sqrt(12) = sqrt(4*3) = sqrt(4)*sqrt(3) = 2sqrt(3)}}}
{{{x = 2sqrt(3)}}} or {{{x = -2sqrt(3)}}}