Solve 9+left(1+xright)^2={left(3+sqrt{x^2-1}right)}^2

$$9+ { \left(1+x \right) }^{ 2 } = { \left(3+ \sqrt{ { x }^{ 2 } -1 } \right) }^{ 2 }$$

Solve for x

$$x=-1$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Solution Steps

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

$$9+1+2x+x^{2}=\left(3+\sqrt{x^{2}-1}\right)^{2}$$

Add 9 and 1 to get 10.

$$10+2x+x^{2}=\left(3+\sqrt{x^{2}-1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{x^{2}-1}\right)^{2}.

$$10+2x+x^{2}=9+6\sqrt{x^{2}-1}+\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

$$10+2x+x^{2}=9+6\sqrt{x^{2}-1}+x^{2}-1$$

Subtract 1 from 9 to get 8.

$$10+2x+x^{2}=8+6\sqrt{x^{2}-1}+x^{2}$$

Subtract 6\sqrt{x^{2}-1} from both sides.

$$10+2x+x^{2}-6\sqrt{x^{2}-1}=8+x^{2}$$

Subtract x^{2} from both sides.

$$10+2x+x^{2}-6\sqrt{x^{2}-1}-x^{2}=8$$

Combine x^{2} and -x^{2} to get 0.

$$10+2x-6\sqrt{x^{2}-1}=8$$

Subtract 10+2x from both sides of the equation.

$$-6\sqrt{x^{2}-1}=8-\left(10+2x\right)$$

To find the opposite of 10+2x, find the opposite of each term.

$$-6\sqrt{x^{2}-1}=8-10-2x$$

Subtract 10 from 8 to get -2.

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

Square both sides of the equation.

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

Expand \left(-6\sqrt{x^{2}-1}\right)^{2}.

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

Calculate -6 to the power of 2 and get 36.

$$36\left(\sqrt{x^{2}-1}\right)^{2}=\left(-2-2x\right)^{2}$$

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

$$36\left(x^{2}-1\right)=\left(-2-2x\right)^{2}$$

Use the distributive property to multiply 36 by x^{2}-1.

$$36x^{2}-36=\left(-2-2x\right)^{2}$$

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

$$36x^{2}-36=4+8x+4x^{2}$$

Subtract 4 from both sides.

$$36x^{2}-36-4=8x+4x^{2}$$

Subtract 4 from -36 to get -40.

$$36x^{2}-40=8x+4x^{2}$$

Subtract 8x from both sides.

$$36x^{2}-40-8x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$36x^{2}-40-8x-4x^{2}=0$$

Combine 36x^{2} and -4x^{2} to get 32x^{2}.

$$32x^{2}-40-8x=0$$

Divide both sides by 8.

$$4x^{2}-5-x=0$$

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

$$4x^{2}-x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-1 ab=4\left(-5\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite 4x^{2}-x-5 as \left(4x^{2}-5x\right)+\left(4x-5\right).

$$\left(4x^{2}-5x\right)+\left(4x-5\right)$$

Factor out x in 4x^{2}-5x.

$$x\left(4x-5\right)+4x-5$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(x+1\right)$$

To find equation solutions, solve 4x-5=0 and x+1=0.

$$x=\frac{5}{4} x=-1$$

Substitute \frac{5}{4} for x in the equation 9+\left(1+x\right)^{2}=\left(3+\sqrt{x^{2}-1}\right)^{2}.

$$9+\left(1+\frac{5}{4}\right)^{2}=\left(3+\sqrt{\left(\frac{5}{4}\right)^{2}-1}\right)^{2}$$

Simplify. The value x=\frac{5}{4} satisfies the equation.

$$\frac{225}{16}=\frac{225}{16}$$

Substitute -1 for x in the equation 9+\left(1+x\right)^{2}=\left(3+\sqrt{x^{2}-1}\right)^{2}.

$$9+\left(1-1\right)^{2}=\left(3+\sqrt{\left(-1\right)^{2}-1}\right)^{2}$$

Simplify. The value x=-1 satisfies the equation.

$$9=9$$

List all solutions of -6\sqrt{x^{2}-1}=-2x-2.

$$x=\frac{5}{4} x=-1$$