Solve sqrt{x+7}+sqrt{x+2}=sqrt{6x+13}

$$\sqrt { x + 7 } + \sqrt { x + 2 } = \sqrt { 6 x + 13 }$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+7}+\sqrt{x+2}\right)^{2}=\left(\sqrt{6x+13}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+7}+\sqrt{x+2}\right)^{2}.

$$\left(\sqrt{x+7}\right)^{2}+2\sqrt{x+7}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}=\left(\sqrt{6x+13}\right)^{2}$$

Calculate \sqrt{x+7} to the power of 2 and get x+7.

$$x+7+2\sqrt{x+7}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}=\left(\sqrt{6x+13}\right)^{2}$$

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

$$x+7+2\sqrt{x+7}\sqrt{x+2}+x+2=\left(\sqrt{6x+13}\right)^{2}$$

Combine x and x to get 2x.

$$2x+7+2\sqrt{x+7}\sqrt{x+2}+2=\left(\sqrt{6x+13}\right)^{2}$$

Add 7 and 2 to get 9.

$$2x+9+2\sqrt{x+7}\sqrt{x+2}=\left(\sqrt{6x+13}\right)^{2}$$

Calculate \sqrt{6x+13} to the power of 2 and get 6x+13.

$$2x+9+2\sqrt{x+7}\sqrt{x+2}=6x+13$$

Subtract 2x+9 from both sides of the equation.

$$2\sqrt{x+7}\sqrt{x+2}=6x+13-\left(2x+9\right)$$

To find the opposite of 2x+9, find the opposite of each term.

$$2\sqrt{x+7}\sqrt{x+2}=6x+13-2x-9$$

Combine 6x and -2x to get 4x.

$$2\sqrt{x+7}\sqrt{x+2}=4x+13-9$$

Subtract 9 from 13 to get 4.

$$2\sqrt{x+7}\sqrt{x+2}=4x+4$$

Square both sides of the equation.

$$\left(2\sqrt{x+7}\sqrt{x+2}\right)^{2}=\left(4x+4\right)^{2}$$

Expand \left(2\sqrt{x+7}\sqrt{x+2}\right)^{2}.

$$2^{2}\left(\sqrt{x+7}\right)^{2}\left(\sqrt{x+2}\right)^{2}=\left(4x+4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+7}\right)^{2}\left(\sqrt{x+2}\right)^{2}=\left(4x+4\right)^{2}$$

Calculate \sqrt{x+7} to the power of 2 and get x+7.

$$4\left(x+7\right)\left(\sqrt{x+2}\right)^{2}=\left(4x+4\right)^{2}$$

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

$$4\left(x+7\right)\left(x+2\right)=\left(4x+4\right)^{2}$$

Use the distributive property to multiply 4 by x+7.

$$\left(4x+28\right)\left(x+2\right)=\left(4x+4\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+28 by each term of x+2.

$$4x^{2}+8x+28x+56=\left(4x+4\right)^{2}$$

Combine 8x and 28x to get 36x.

$$4x^{2}+36x+56=\left(4x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+4\right)^{2}.

$$4x^{2}+36x+56=16x^{2}+32x+16$$

Subtract 16x^{2} from both sides.

$$4x^{2}+36x+56-16x^{2}=32x+16$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}+36x+56=32x+16$$

Subtract 32x from both sides.

$$-12x^{2}+36x+56-32x=16$$

Combine 36x and -32x to get 4x.

$$-12x^{2}+4x+56=16$$

Subtract 16 from both sides.

$$-12x^{2}+4x+56-16=0$$

Subtract 16 from 56 to get 40.

$$-12x^{2}+4x+40=0$$

Divide both sides by 4.

$$-3x^{2}+x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=1 ab=-3\times 10=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=6 b=-5$$

Rewrite -3x^{2}+x+10 as \left(-3x^{2}+6x\right)+\left(-5x+10\right).

$$\left(-3x^{2}+6x\right)+\left(-5x+10\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(-x+2\right)+5\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(3x+5\right)$$

To find equation solutions, solve -x+2=0 and 3x+5=0.

$$x=2 x=-\frac{5}{3}$$

Substitute 2 for x in the equation \sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}.

$$\sqrt{2+7}+\sqrt{2+2}=\sqrt{6\times 2+13}$$

Simplify. The value x=2 satisfies the equation.

$$5=5$$

Substitute -\frac{5}{3} for x in the equation \sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}.

$$\sqrt{-\frac{5}{3}+7}+\sqrt{-\frac{5}{3}+2}=\sqrt{6\left(-\frac{5}{3}\right)+13}$$

Simplify. The value x=-\frac{5}{3} does not satisfy the equation.

$$\frac{5}{3}\times 3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

Substitute 2 for x in the equation \sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}.

$$\sqrt{2+7}+\sqrt{2+2}=\sqrt{6\times 2+13}$$

Simplify. The value x=2 satisfies the equation.

$$5=5$$

Equation \sqrt{x+2}+\sqrt{x+7}=\sqrt{6x+13} has a unique solution.

$$x=2$$