Solve sqrt{4x}+5=sqrt{x+32}

$$\sqrt { 4 x } + 5 = \sqrt { x + 32 }$$

Solve for x

$$x=\frac{1}{9}\approx 0.111111111$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x}+5\right)^{2}=\left(\sqrt{x+32}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{4x}+5\right)^{2}.

$$\left(\sqrt{4x}\right)^{2}+10\sqrt{4x}+25=\left(\sqrt{x+32}\right)^{2}$$

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

$$4x+10\sqrt{4x}+25=\left(\sqrt{x+32}\right)^{2}$$

Calculate \sqrt{x+32} to the power of 2 and get x+32.

$$4x+10\sqrt{4x}+25=x+32$$

Subtract 4x+25 from both sides of the equation.

$$10\sqrt{4x}=x+32-\left(4x+25\right)$$

To find the opposite of 4x+25, find the opposite of each term.

$$10\sqrt{4x}=x+32-4x-25$$

Combine x and -4x to get -3x.

$$10\sqrt{4x}=-3x+32-25$$

Subtract 25 from 32 to get 7.

$$10\sqrt{4x}=-3x+7$$

Square both sides of the equation.

$$\left(10\sqrt{4x}\right)^{2}=\left(-3x+7\right)^{2}$$

Expand \left(10\sqrt{4x}\right)^{2}.

$$10^{2}\left(\sqrt{4x}\right)^{2}=\left(-3x+7\right)^{2}$$

Calculate 10 to the power of 2 and get 100.

$$100\left(\sqrt{4x}\right)^{2}=\left(-3x+7\right)^{2}$$

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

$$100\times 4x=\left(-3x+7\right)^{2}$$

Multiply 100 and 4 to get 400.

$$400x=\left(-3x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-3x+7\right)^{2}.

$$400x=9x^{2}-42x+49$$

Subtract 9x^{2} from both sides.

$$400x-9x^{2}=-42x+49$$

Add 42x to both sides.

$$400x-9x^{2}+42x=49$$

Combine 400x and 42x to get 442x.

$$442x-9x^{2}=49$$

Subtract 49 from both sides.

$$442x-9x^{2}-49=0$$

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

$$-9x^{2}+442x-49=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-49. To find a and b, set up a system to be solved.

$$a+b=442 ab=-9\left(-49\right)=441$$

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 441.

$$1,441 3,147 7,63 9,49 21,21$$

Calculate the sum for each pair.

$$1+441=442 3+147=150 7+63=70 9+49=58 21+21=42$$

The solution is the pair that gives sum 442.

$$a=441 b=1$$

Rewrite -9x^{2}+442x-49 as \left(-9x^{2}+441x\right)+\left(x-49\right).

$$\left(-9x^{2}+441x\right)+\left(x-49\right)$$

Factor out 9x in the first and -1 in the second group.

$$9x\left(-x+49\right)-\left(-x+49\right)$$

Factor out common term -x+49 by using distributive property.

$$\left(-x+49\right)\left(9x-1\right)$$

To find equation solutions, solve -x+49=0 and 9x-1=0.

$$x=49 x=\frac{1}{9}$$

Substitute 49 for x in the equation \sqrt{4x}+5=\sqrt{x+32}.

$$\sqrt{4\times 49}+5=\sqrt{49+32}$$

Simplify. The value x=49 does not satisfy the equation.

$$19=9$$

Substitute \frac{1}{9} for x in the equation \sqrt{4x}+5=\sqrt{x+32}.

$$\sqrt{4\times \frac{1}{9}}+5=\sqrt{\frac{1}{9}+32}$$

Simplify. The value x=\frac{1}{9} satisfies the equation.

$$\frac{17}{3}=\frac{17}{3}$$

Equation \sqrt{4x}+5=\sqrt{x+32} has a unique solution.

$$x=\frac{1}{9}$$