Solve sqrt{x^2+3}=4x-2

$$\sqrt { x ^ { 2 } + 3 } = 4 x - 2$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+3}\right)^{2}=\left(4x-2\right)^{2}$$

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

$$x^{2}+3=\left(4x-2\right)^{2}$$

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

$$x^{2}+3=16x^{2}-16x+4$$

Subtract 16x^{2} from both sides.

$$x^{2}+3-16x^{2}=-16x+4$$

Combine x^{2} and -16x^{2} to get -15x^{2}.

$$-15x^{2}+3=-16x+4$$

Add 16x to both sides.

$$-15x^{2}+3+16x=4$$

Subtract 4 from both sides.

$$-15x^{2}+3+16x-4=0$$

Subtract 4 from 3 to get -1.

$$-15x^{2}-1+16x=0$$

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

$$-15x^{2}+16x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -15x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=16 ab=-15\left(-1\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 16.

$$a=15 b=1$$

Rewrite -15x^{2}+16x-1 as \left(-15x^{2}+15x\right)+\left(x-1\right).

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

Factor out 15x in the first and -1 in the second group.

$$15x\left(-x+1\right)-\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(15x-1\right)$$

To find equation solutions, solve -x+1=0 and 15x-1=0.

$$x=1 x=\frac{1}{15}$$

Substitute 1 for x in the equation \sqrt{x^{2}+3}=4x-2.

$$\sqrt{1^{2}+3}=4\times 1-2$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Substitute \frac{1}{15} for x in the equation \sqrt{x^{2}+3}=4x-2.

$$\sqrt{\left(\frac{1}{15}\right)^{2}+3}=4\times \frac{1}{15}-2$$

Simplify. The value x=\frac{1}{15} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{26}{15}=-\frac{26}{15}$$

Equation \sqrt{x^{2}+3}=4x-2 has a unique solution.

$$x=1$$