Solve (x-10)=(3sqrt{x})

$$(x-10)=(3 \sqrt{ x } )$$

Solve for x

$$x=25$$

Solution Steps

Square both sides of the equation.

$$\left(x-10\right)^{2}=\left(3\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-10\right)^{2}.

$$x^{2}-20x+100=\left(3\sqrt{x}\right)^{2}$$

Expand \left(3\sqrt{x}\right)^{2}.

$$x^{2}-20x+100=3^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-20x+100=9\left(\sqrt{x}\right)^{2}$$

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

$$x^{2}-20x+100=9x$$

Subtract 9x from both sides.

$$x^{2}-20x+100-9x=0$$

Combine -20x and -9x to get -29x.

$$x^{2}-29x+100=0$$

To solve the equation, factor x^{2}-29x+100 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.

$$a+b=-29 ab=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -29.

$$a=-25 b=-4$$

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

$$\left(x-25\right)\left(x-4\right)$$

To find equation solutions, solve x-25=0 and x-4=0.

$$x=25 x=4$$

Substitute 25 for x in the equation x-10=3\sqrt{x}.

$$25-10=3\sqrt{25}$$

Simplify. The value x=25 satisfies the equation.

$$15=15$$

Substitute 4 for x in the equation x-10=3\sqrt{x}.

$$4-10=3\sqrt{4}$$

Simplify. The value x=4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-6=6$$

Equation x-10=3\sqrt{x} has a unique solution.

$$x=25$$