Solve sqrt{6+sqrt{x+4}}=sqrt{2x-1}

$$\sqrt{ 6+ \sqrt{ x+4 } } = \sqrt{ 2x-1 }$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{6+\sqrt{x+4}} to the power of 2 and get 6+\sqrt{x+4}.

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

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

$$6+\sqrt{x+4}=2x-1$$

Subtract 6 from both sides of the equation.

$$\sqrt{x+4}=2x-1-6$$

Subtract 6 from -1 to get -7.

$$\sqrt{x+4}=2x-7$$

Square both sides of the equation.

$$\left(\sqrt{x+4}\right)^{2}=\left(2x-7\right)^{2}$$

Calculate \sqrt{x+4} to the power of 2 and get x+4.

$$x+4=\left(2x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-7\right)^{2}.

$$x+4=4x^{2}-28x+49$$

Subtract 4x^{2} from both sides.

$$x+4-4x^{2}=-28x+49$$

Add 28x to both sides.

$$x+4-4x^{2}+28x=49$$

Combine x and 28x to get 29x.

$$29x+4-4x^{2}=49$$

Subtract 49 from both sides.

$$29x+4-4x^{2}-49=0$$

Subtract 49 from 4 to get -45.

$$29x-45-4x^{2}=0$$

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

$$-4x^{2}+29x-45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=29 ab=-4\left(-45\right)=180$$

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 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 29.

$$a=20 b=9$$

Rewrite -4x^{2}+29x-45 as \left(-4x^{2}+20x\right)+\left(9x-45\right).

$$\left(-4x^{2}+20x\right)+\left(9x-45\right)$$

Factor out 4x in the first and -9 in the second group.

$$4x\left(-x+5\right)-9\left(-x+5\right)$$

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

$$\left(-x+5\right)\left(4x-9\right)$$

To find equation solutions, solve -x+5=0 and 4x-9=0.

$$x=5 x=\frac{9}{4}$$

Substitute 5 for x in the equation \sqrt{6+\sqrt{x+4}}=\sqrt{2x-1}.

$$\sqrt{6+\sqrt{5+4}}=\sqrt{2\times 5-1}$$

Simplify. The value x=5 satisfies the equation.

$$3=3$$

Substitute \frac{9}{4} for x in the equation \sqrt{6+\sqrt{x+4}}=\sqrt{2x-1}.

$$\sqrt{6+\sqrt{\frac{9}{4}+4}}=\sqrt{2\times \frac{9}{4}-1}$$

Simplify. The value x=\frac{9}{4} does not satisfy the equation.

$$\frac{1}{2}\times 34^{\frac{1}{2}}=\frac{1}{2}\times 14^{\frac{1}{2}}$$

Substitute 5 for x in the equation \sqrt{6+\sqrt{x+4}}=\sqrt{2x-1}.

$$\sqrt{6+\sqrt{5+4}}=\sqrt{2\times 5-1}$$

Simplify. The value x=5 satisfies the equation.

$$3=3$$

Equation \sqrt{\sqrt{x+4}+6}=\sqrt{2x-1} has a unique solution.

$$x=5$$