Question 417839
{{{4x^2-28x+49=5}}}
Solving equations this way involves the following steps:<ol><li>Gather the variable terms on one side of the equation.</li><li>Make a perfect square out the side of the equation with the variables.</li><li>Find the square root of each side of the equation. This results in two equations! One for the positive square root and one for the negative square root.)</li><li>Solve the remaining equations.</li></ol>Let's see how this works:
1) Gather...
The variable terms are already just on the left side of the equation.
2) Make a perfect square...
The left side, where the variable terms are, is already a perfect square! It fits the pattern for
{{{a^2-2ab+b^2 = (a-b)^2}}}
with the "a" being 2x and the "b" being 7. So we can rewrite the left side as a perfect square:
{{{(2x-7)^2 = 5}}}
3) Find the square root of each side.
{{{sqrt((2x-7)^2) = sqrt(5)}}}
which simplifies to:
{{{abs(2x-7) = sqrt(5)}}}
which becomes:
{{{2x-7 = sqrt(5)}}} or {{{2x-7 = -sqrt(5)}}}
4) Solve the equations.
Adding 7 to each side:
{{{2x = 7 + sqrt(5)}}} or {{{2x = 7 - sqrt(5)}}}
Dividing by 2:
{{{x = (7 + sqrt(5))/2}}} or {{{x = (7 - sqrt(5))/2}}}