Solve x=5sqrt{x}-6

$$x=5 \sqrt{ x } -6$$

Solve for x

$$x=9$$

$$x=4$$

Solution Steps

Subtract -6 from both sides of the equation.

$$x+6=5\sqrt{x}$$

Square both sides of the equation.

$$\left(x+6\right)^{2}=\left(5\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+12x+36=\left(5\sqrt{x}\right)^{2}$$

Expand \left(5\sqrt{x}\right)^{2}.

$$x^{2}+12x+36=5^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$x^{2}+12x+36=25\left(\sqrt{x}\right)^{2}$$

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+12x+36=25x$$

Subtract 25x from both sides.

$$x^{2}+12x+36-25x=0$$

Combine 12x and -25x to get -13x.

$$x^{2}-13x+36=0$$

To solve the equation, factor x^{2}-13x+36 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-9\right)\left(x-4\right)$$

To find equation solutions, solve x-9=0 and x-4=0.

$$x=9 x=4$$

Substitute 9 for x in the equation x=5\sqrt{x}-6.

$$9=5\sqrt{9}-6$$

Simplify. The value x=9 satisfies the equation.

$$9=9$$

Substitute 4 for x in the equation x=5\sqrt{x}-6.

$$4=5\sqrt{4}-6$$

Simplify. The value x=4 satisfies the equation.

$$4=4$$

List all solutions of x+6=5\sqrt{x}.

$$x=9 x=4$$