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

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

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.

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