Solve sqrt{3x+10}-5=-3sqrt{x+3}

$$\sqrt{ 3x+10 } -5 = -3 \sqrt{ x+3 }$$

Solve for x

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+10}-5\right)^{2}=\left(-3\sqrt{x+3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3x+10}-5\right)^{2}.

$$\left(\sqrt{3x+10}\right)^{2}-10\sqrt{3x+10}+25=\left(-3\sqrt{x+3}\right)^{2}$$

Calculate \sqrt{3x+10} to the power of 2 and get 3x+10.

$$3x+10-10\sqrt{3x+10}+25=\left(-3\sqrt{x+3}\right)^{2}$$

Add 10 and 25 to get 35.

$$3x+35-10\sqrt{3x+10}=\left(-3\sqrt{x+3}\right)^{2}$$

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

$$3x+35-10\sqrt{3x+10}=\left(-3\right)^{2}\left(\sqrt{x+3}\right)^{2}$$

Calculate -3 to the power of 2 and get 9.

$$3x+35-10\sqrt{3x+10}=9\left(\sqrt{x+3}\right)^{2}$$

Calculate \sqrt{x+3} to the power of 2 and get x+3.

$$3x+35-10\sqrt{3x+10}=9\left(x+3\right)$$

Use the distributive property to multiply 9 by x+3.

$$3x+35-10\sqrt{3x+10}=9x+27$$

Subtract 3x+35 from both sides of the equation.

$$-10\sqrt{3x+10}=9x+27-\left(3x+35\right)$$

To find the opposite of 3x+35, find the opposite of each term.

$$-10\sqrt{3x+10}=9x+27-3x-35$$

Combine 9x and -3x to get 6x.

$$-10\sqrt{3x+10}=6x+27-35$$

Subtract 35 from 27 to get -8.

$$-10\sqrt{3x+10}=6x-8$$

Square both sides of the equation.

$$\left(-10\sqrt{3x+10}\right)^{2}=\left(6x-8\right)^{2}$$

Expand \left(-10\sqrt{3x+10}\right)^{2}.

$$\left(-10\right)^{2}\left(\sqrt{3x+10}\right)^{2}=\left(6x-8\right)^{2}$$

Calculate -10 to the power of 2 and get 100.

$$100\left(\sqrt{3x+10}\right)^{2}=\left(6x-8\right)^{2}$$

Calculate \sqrt{3x+10} to the power of 2 and get 3x+10.

$$100\left(3x+10\right)=\left(6x-8\right)^{2}$$

Use the distributive property to multiply 100 by 3x+10.

$$300x+1000=\left(6x-8\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-8\right)^{2}.

$$300x+1000=36x^{2}-96x+64$$

Subtract 36x^{2} from both sides.

$$300x+1000-36x^{2}=-96x+64$$

Add 96x to both sides.

$$300x+1000-36x^{2}+96x=64$$

Combine 300x and 96x to get 396x.

$$396x+1000-36x^{2}=64$$

Subtract 64 from both sides.

$$396x+1000-36x^{2}-64=0$$

Subtract 64 from 1000 to get 936.

$$396x+936-36x^{2}=0$$

Divide both sides by 36.

$$11x+26-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+11x+26=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+26. To find a and b, set up a system to be solved.

$$a+b=11 ab=-26=-26$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -26.

$$-1,26 -2,13$$

Calculate the sum for each pair.

$$-1+26=25 -2+13=11$$

The solution is the pair that gives sum 11.

$$a=13 b=-2$$

Rewrite -x^{2}+11x+26 as \left(-x^{2}+13x\right)+\left(-2x+26\right).

$$\left(-x^{2}+13x\right)+\left(-2x+26\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-13\right)-2\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(-x-2\right)$$

To find equation solutions, solve x-13=0 and -x-2=0.

$$x=13 x=-2$$

Substitute 13 for x in the equation \sqrt{3x+10}-5=-3\sqrt{x+3}.

$$\sqrt{3\times 13+10}-5=-3\sqrt{13+3}$$

Simplify. The value x=13 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-12$$

Substitute -2 for x in the equation \sqrt{3x+10}-5=-3\sqrt{x+3}.

$$\sqrt{3\left(-2\right)+10}-5=-3\sqrt{-2+3}$$

Simplify. The value x=-2 satisfies the equation.

$$-3=-3$$

Equation \sqrt{3x+10}-5=-3\sqrt{x+3} has a unique solution.

$$x=-2$$