Solve sqrt{4x-3}+sqrt{5x+1}=sqrt{15x+4}

$$\sqrt{ 4x-3 } + \sqrt{ 5x+1 } = \sqrt{ 15x+4 }$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x-3}+\sqrt{5x+1}\right)^{2}=\left(\sqrt{15x+4}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{4x-3}+\sqrt{5x+1}\right)^{2}.

$$\left(\sqrt{4x-3}\right)^{2}+2\sqrt{4x-3}\sqrt{5x+1}+\left(\sqrt{5x+1}\right)^{2}=\left(\sqrt{15x+4}\right)^{2}$$

Calculate \sqrt{4x-3} to the power of 2 and get 4x-3.

$$4x-3+2\sqrt{4x-3}\sqrt{5x+1}+\left(\sqrt{5x+1}\right)^{2}=\left(\sqrt{15x+4}\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$4x-3+2\sqrt{4x-3}\sqrt{5x+1}+5x+1=\left(\sqrt{15x+4}\right)^{2}$$

Combine 4x and 5x to get 9x.

$$9x-3+2\sqrt{4x-3}\sqrt{5x+1}+1=\left(\sqrt{15x+4}\right)^{2}$$

Add -3 and 1 to get -2.

$$9x-2+2\sqrt{4x-3}\sqrt{5x+1}=\left(\sqrt{15x+4}\right)^{2}$$

Calculate \sqrt{15x+4} to the power of 2 and get 15x+4.

$$9x-2+2\sqrt{4x-3}\sqrt{5x+1}=15x+4$$

Subtract 9x-2 from both sides of the equation.

$$2\sqrt{4x-3}\sqrt{5x+1}=15x+4-\left(9x-2\right)$$

To find the opposite of 9x-2, find the opposite of each term.

$$2\sqrt{4x-3}\sqrt{5x+1}=15x+4-9x+2$$

Combine 15x and -9x to get 6x.

$$2\sqrt{4x-3}\sqrt{5x+1}=6x+4+2$$

Add 4 and 2 to get 6.

$$2\sqrt{4x-3}\sqrt{5x+1}=6x+6$$

Square both sides of the equation.

$$\left(2\sqrt{4x-3}\sqrt{5x+1}\right)^{2}=\left(6x+6\right)^{2}$$

Expand \left(2\sqrt{4x-3}\sqrt{5x+1}\right)^{2}.

$$2^{2}\left(\sqrt{4x-3}\right)^{2}\left(\sqrt{5x+1}\right)^{2}=\left(6x+6\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{4x-3}\right)^{2}\left(\sqrt{5x+1}\right)^{2}=\left(6x+6\right)^{2}$$

Calculate \sqrt{4x-3} to the power of 2 and get 4x-3.

$$4\left(4x-3\right)\left(\sqrt{5x+1}\right)^{2}=\left(6x+6\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$4\left(4x-3\right)\left(5x+1\right)=\left(6x+6\right)^{2}$$

Use the distributive property to multiply 4 by 4x-3.

$$\left(16x-12\right)\left(5x+1\right)=\left(6x+6\right)^{2}$$

Apply the distributive property by multiplying each term of 16x-12 by each term of 5x+1.

$$80x^{2}+16x-60x-12=\left(6x+6\right)^{2}$$

Combine 16x and -60x to get -44x.

$$80x^{2}-44x-12=\left(6x+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+6\right)^{2}.

$$80x^{2}-44x-12=36x^{2}+72x+36$$

Subtract 36x^{2} from both sides.

$$80x^{2}-44x-12-36x^{2}=72x+36$$

Combine 80x^{2} and -36x^{2} to get 44x^{2}.

$$44x^{2}-44x-12=72x+36$$

Subtract 72x from both sides.

$$44x^{2}-44x-12-72x=36$$

Combine -44x and -72x to get -116x.

$$44x^{2}-116x-12=36$$

Subtract 36 from both sides.

$$44x^{2}-116x-12-36=0$$

Subtract 36 from -12 to get -48.

$$44x^{2}-116x-48=0$$

Divide both sides by 4.

$$11x^{2}-29x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 11x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-29 ab=11\left(-12\right)=-132$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -132.

$$1,-132 2,-66 3,-44 4,-33 6,-22 11,-12$$

Calculate the sum for each pair.

$$1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1$$

The solution is the pair that gives sum -29.

$$a=-33 b=4$$

Rewrite 11x^{2}-29x-12 as \left(11x^{2}-33x\right)+\left(4x-12\right).

$$\left(11x^{2}-33x\right)+\left(4x-12\right)$$

Factor out 11x in the first and 4 in the second group.

$$11x\left(x-3\right)+4\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(11x+4\right)$$

To find equation solutions, solve x-3=0 and 11x+4=0.

$$x=3 x=-\frac{4}{11}$$

Substitute -\frac{4}{11} for x in the equation \sqrt{4x-3}+\sqrt{5x+1}=\sqrt{15x+4}. The expression \sqrt{4\left(-\frac{4}{11}\right)-3} is undefined because the radicand cannot be negative.

$$\sqrt{4\left(-\frac{4}{11}\right)-3}+\sqrt{5\left(-\frac{4}{11}\right)+1}=\sqrt{15\left(-\frac{4}{11}\right)+4}$$

Substitute 3 for x in the equation \sqrt{4x-3}+\sqrt{5x+1}=\sqrt{15x+4}.

$$\sqrt{4\times 3-3}+\sqrt{5\times 3+1}=\sqrt{15\times 3+4}$$

Simplify. The value x=3 satisfies the equation.

$$7=7$$

Equation \sqrt{5x+1}+\sqrt{4x-3}=\sqrt{15x+4} has a unique solution.

$$x=3$$