Question 70971
#30. Solve:
{{{sqrt(2y+7)+4 = y}}} Subtract 4 from both sides of the equation.
{{{sqrt(2y+7) = y-4}}} Now square both sides.
{{{2y+7 = (y-4)^2}}} Simplify the right side.
{{{2y+7 = y^2-8y+16}}} Subtract (2y+7) from both sides.
{{{y^2-10y+9 = 0}}} Factor the quadratic equation.
{{{(y-1)(y-9) = 0}}} Apply the zero product principle.
{{{y-1 = 0}}} and/or {{{y-9 = 0}}}
{{{y = 1}}} and {{{y = 9}}}
Check:
y = 1
{{{sqrt(2(1)+7) + 4 = 1}}}
{{{sqrt(9)+4 = 1}}} Note that {{{sqrt(9) = -3}}} or +3
{{{-3+4 = 1}}} The extraneous root of (+3) was introduced when squaring {{{sqrt(2y+7)}}}
y= 9
{{{sqrt(2(9)+7) + 4 = 9}}}
{{{sqrt(25) + 4 =9}}} Note that {{{sqrt(25) = 5}}} or -5
{{{5+4 = 9}}} The extaneous root of (-5) was introduced when squaring {{{sqrt(2y+7)}}}

#16
The geometry of the problem describes a right triangle in which the height is 3 m. and the length of the hypotenuse is 7 m.
Use the Pythagorean theorem {{{c^2 = a^2+b^2}}} to calculate the third side of this triangle. The hypotenuse is 7 m. so c = 7 and the height is 3 m. so a = 3
{{{7^2 = 3^2 + b^2}}} Solve for b which is the required length of the insulation.
{{{49 = 9 + b^2}}} Subtract 9 from both sides.
{{{40 = b^2}}} Take the square root of both sides.
{{{sqrt(40) = b}}}
{{{b = 6.3m}}} To the nearest tenth.