Solve 1sqrt{x^2+3x-3}-sqrt{6x+7}=0

$$1 \sqrt{ { x }^{ 2 } +3x-3 } - \sqrt{ 6x+7 } = 0$$

Solve for x (complex solution)

$$x=-2$$

$$x=5$$

Solution Steps

Subtract -\sqrt{6x+7} from both sides of the equation.

$$\sqrt{x^{2}+3x-3}=\sqrt{6x+7}$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+3x-3}\right)^{2}=\left(\sqrt{6x+7}\right)^{2}$$

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

$$x^{2}+3x-3=\left(\sqrt{6x+7}\right)^{2}$$

Calculate \sqrt{6x+7} to the power of 2 and get 6x+7.

$$x^{2}+3x-3=6x+7$$

Subtract 6x from both sides.

$$x^{2}+3x-3-6x=7$$

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

$$x^{2}-3x-3=7$$

Subtract 7 from both sides.

$$x^{2}-3x-3-7=0$$

Subtract 7 from -3 to get -10.

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

To solve the equation, factor x^{2}-3x-10 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.

$$a+b=-3 ab=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

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

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

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

$$x=5 x=-2$$

Substitute 5 for x in the equation 1\sqrt{x^{2}+3x-3}-\sqrt{6x+7}=0.

$$1\sqrt{5^{2}+3\times 5-3}-\sqrt{6\times 5+7}=0$$

Simplify. The value x=5 satisfies the equation.

$$0=0$$

Substitute -2 for x in the equation 1\sqrt{x^{2}+3x-3}-\sqrt{6x+7}=0.

$$1\sqrt{\left(-2\right)^{2}+3\left(-2\right)-3}-\sqrt{6\left(-2\right)+7}=0$$

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

$$0=0$$

List all solutions of \sqrt{x^{2}+3x-3}=\sqrt{6x+7}.

$$x=5 x=-2$$

Solve for x

$$x=5$$

Solution Steps

Subtract -\sqrt{6x+7} from both sides of the equation.

$$\sqrt{x^{2}+3x-3}=\sqrt{6x+7}$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+3x-3}\right)^{2}=\left(\sqrt{6x+7}\right)^{2}$$

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

$$x^{2}+3x-3=\left(\sqrt{6x+7}\right)^{2}$$

Calculate \sqrt{6x+7} to the power of 2 and get 6x+7.

$$x^{2}+3x-3=6x+7$$

Subtract 6x from both sides.

$$x^{2}+3x-3-6x=7$$

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

$$x^{2}-3x-3=7$$

Subtract 7 from both sides.

$$x^{2}-3x-3-7=0$$

Subtract 7 from -3 to get -10.

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

To solve the equation, factor x^{2}-3x-10 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.

$$a+b=-3 ab=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

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

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

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

$$x=5 x=-2$$

Substitute 5 for x in the equation 1\sqrt{x^{2}+3x-3}-\sqrt{6x+7}=0.

$$1\sqrt{5^{2}+3\times 5-3}-\sqrt{6\times 5+7}=0$$

Simplify. The value x=5 satisfies the equation.

$$0=0$$

Substitute -2 for x in the equation 1\sqrt{x^{2}+3x-3}-\sqrt{6x+7}=0. The expression \sqrt{\left(-2\right)^{2}+3\left(-2\right)-3} is undefined because the radicand cannot be negative.

$$1\sqrt{\left(-2\right)^{2}+3\left(-2\right)-3}-\sqrt{6\left(-2\right)+7}=0$$

Equation \sqrt{x^{2}+3x-3}=\sqrt{6x+7} has a unique solution.

$$x=5$$