Solve sqrt{4x+2}=sqrt{2sqrt{-5x+29}}

$$\sqrt{ 4x+2 } = \sqrt{ 2 \sqrt{ -5x+29 } }$$

Solve for x

$$x = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x+2}\right)^{2}=\left(\sqrt{2\sqrt{-5x+29}}\right)^{2}$$

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

$$4x+2=\left(\sqrt{2\sqrt{-5x+29}}\right)^{2}$$

Calculate \sqrt{2\sqrt{-5x+29}} to the power of 2 and get 2\sqrt{-5x+29}.

$$4x+2=2\sqrt{-5x+29}$$

Square both sides of the equation.

$$\left(4x+2\right)^{2}=\left(2\sqrt{-5x+29}\right)^{2}$$

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

$$16x^{2}+16x+4=\left(2\sqrt{-5x+29}\right)^{2}$$

Expand \left(2\sqrt{-5x+29}\right)^{2}.

$$16x^{2}+16x+4=2^{2}\left(\sqrt{-5x+29}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$16x^{2}+16x+4=4\left(\sqrt{-5x+29}\right)^{2}$$

Calculate \sqrt{-5x+29} to the power of 2 and get -5x+29.

$$16x^{2}+16x+4=4\left(-5x+29\right)$$

Use the distributive property to multiply 4 by -5x+29.

$$16x^{2}+16x+4=-20x+116$$

Add 20x to both sides.

$$16x^{2}+16x+4+20x=116$$

Combine 16x and 20x to get 36x.

$$16x^{2}+36x+4=116$$

Subtract 116 from both sides.

$$16x^{2}+36x+4-116=0$$

Subtract 116 from 4 to get -112.

$$16x^{2}+36x-112=0$$

Divide both sides by 4.

$$4x^{2}+9x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=9 ab=4\left(-28\right)=-112$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -112.

$$-1,112 -2,56 -4,28 -7,16 -8,14$$

Calculate the sum for each pair.

$$-1+112=111 -2+56=54 -4+28=24 -7+16=9 -8+14=6$$

The solution is the pair that gives sum 9.

$$a=-7 b=16$$

Rewrite 4x^{2}+9x-28 as \left(4x^{2}-7x\right)+\left(16x-28\right).

$$\left(4x^{2}-7x\right)+\left(16x-28\right)$$

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

$$x\left(4x-7\right)+4\left(4x-7\right)$$

Factor out common term 4x-7 by using distributive property.

$$\left(4x-7\right)\left(x+4\right)$$

To find equation solutions, solve 4x-7=0 and x+4=0.

$$x=\frac{7}{4} x=-4$$

Substitute -4 for x in the equation \sqrt{4x+2}=\sqrt{2\sqrt{-5x+29}}. The expression \sqrt{4\left(-4\right)+2} is undefined because the radicand cannot be negative.

$$\sqrt{4\left(-4\right)+2}=\sqrt{2\sqrt{-5\left(-4\right)+29}}$$

Substitute \frac{7}{4} for x in the equation \sqrt{4x+2}=\sqrt{2\sqrt{-5x+29}}.

$$\sqrt{4\times \frac{7}{4}+2}=\sqrt{2\sqrt{-5\times \frac{7}{4}+29}}$$

Simplify. The value x=\frac{7}{4} satisfies the equation.

$$3=3$$

Equation \sqrt{4x+2}=\sqrt{2\sqrt{29-5x}} has a unique solution.

$$x=\frac{7}{4}$$