Solve sqrt{6x^2-15}=x+1

$$\sqrt { 6 x ^ { 2 } - 15 } = x + 1$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{6x^{2}-15}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{6x^{2}-15} to the power of 2 and get 6x^{2}-15.

$$6x^{2}-15=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$6x^{2}-15=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$6x^{2}-15-x^{2}=2x+1$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-15=2x+1$$

Subtract 2x from both sides.

$$5x^{2}-15-2x=1$$

Subtract 1 from both sides.

$$5x^{2}-15-2x-1=0$$

Subtract 1 from -15 to get -16.

$$5x^{2}-16-2x=0$$

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

$$5x^{2}-2x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-2 ab=5\left(-16\right)=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -2.

$$a=-10 b=8$$

Rewrite 5x^{2}-2x-16 as \left(5x^{2}-10x\right)+\left(8x-16\right).

$$\left(5x^{2}-10x\right)+\left(8x-16\right)$$

Factor out 5x in the first and 8 in the second group.

$$5x\left(x-2\right)+8\left(x-2\right)$$

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

$$\left(x-2\right)\left(5x+8\right)$$

To find equation solutions, solve x-2=0 and 5x+8=0.

$$x=2 x=-\frac{8}{5}$$

Substitute 2 for x in the equation \sqrt{6x^{2}-15}=x+1.

$$\sqrt{6\times 2^{2}-15}=2+1$$

Simplify. The value x=2 satisfies the equation.

$$3=3$$

Substitute -\frac{8}{5} for x in the equation \sqrt{6x^{2}-15}=x+1.

$$\sqrt{6\left(-\frac{8}{5}\right)^{2}-15}=-\frac{8}{5}+1$$

Simplify. The value x=-\frac{8}{5} does not satisfy the equation because the left and the right hand side have opposite signs.

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

Equation \sqrt{6x^{2}-15}=x+1 has a unique solution.

$$x=2$$