Solve 3x=sqrt{x^2+6}-4

$$3x = \sqrt{ { x }^{ 2 } +6 } -4$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Solution Steps

Subtract -4 from both sides of the equation.

$$3x+4=\sqrt{x^{2}+6}$$

Square both sides of the equation.

$$\left(3x+4\right)^{2}=\left(\sqrt{x^{2}+6}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$9x^{2}+24x+16=\left(\sqrt{x^{2}+6}\right)^{2}$$

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

$$9x^{2}+24x+16=x^{2}+6$$

Subtract x^{2} from both sides.

$$9x^{2}+24x+16-x^{2}=6$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}+24x+16=6$$

Subtract 6 from both sides.

$$8x^{2}+24x+16-6=0$$

Subtract 6 from 16 to get 10.

$$8x^{2}+24x+10=0$$

Divide both sides by 2.

$$4x^{2}+12x+5=0$$

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.

$$a+b=12 ab=4\times 5=20$$

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 20.

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

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 12.

$$a=2 b=10$$

Rewrite 4x^{2}+12x+5 as \left(4x^{2}+2x\right)+\left(10x+5\right).

$$\left(4x^{2}+2x\right)+\left(10x+5\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(2x+1\right)+5\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(2x+5\right)$$

To find equation solutions, solve 2x+1=0 and 2x+5=0.

$$x=-\frac{1}{2} x=-\frac{5}{2}$$

Substitute -\frac{1}{2} for x in the equation 3x=\sqrt{x^{2}+6}-4.

$$3\left(-\frac{1}{2}\right)=\sqrt{\left(-\frac{1}{2}\right)^{2}+6}-4$$

Simplify. The value x=-\frac{1}{2} satisfies the equation.

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

Substitute -\frac{5}{2} for x in the equation 3x=\sqrt{x^{2}+6}-4.

$$3\left(-\frac{5}{2}\right)=\sqrt{\left(-\frac{5}{2}\right)^{2}+6}-4$$

Simplify. The value x=-\frac{5}{2} does not satisfy the equation.

$$-\frac{15}{2}=-\frac{1}{2}$$

Equation 3x+4=\sqrt{x^{2}+6} has a unique solution.

$$x=-\frac{1}{2}$$