Solve sqrt{2x+3}-sqrt{x-1}=sqrt{3x-8}

$$\sqrt { 2 x + 3 } - \sqrt { x - 1 } = \sqrt { 3 x - 8 }$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Solution Steps

Subtract -\sqrt{x-1} from both sides of the equation.

$$\sqrt{2x+3}=\sqrt{3x-8}+\sqrt{x-1}$$

Square both sides of the equation.

$$\left(\sqrt{2x+3}\right)^{2}=\left(\sqrt{3x-8}+\sqrt{x-1}\right)^{2}$$

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

$$2x+3=\left(\sqrt{3x-8}+\sqrt{x-1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3x-8}+\sqrt{x-1}\right)^{2}.

$$2x+3=\left(\sqrt{3x-8}\right)^{2}+2\sqrt{3x-8}\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

Calculate \sqrt{3x-8} to the power of 2 and get 3x-8.

$$2x+3=3x-8+2\sqrt{3x-8}\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

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

$$2x+3=3x-8+2\sqrt{3x-8}\sqrt{x-1}+x-1$$

Combine 3x and x to get 4x.

$$2x+3=4x-8+2\sqrt{3x-8}\sqrt{x-1}-1$$

Subtract 1 from -8 to get -9.

$$2x+3=4x-9+2\sqrt{3x-8}\sqrt{x-1}$$

Subtract 4x-9 from both sides of the equation.

$$2x+3-\left(4x-9\right)=2\sqrt{3x-8}\sqrt{x-1}$$

To find the opposite of 4x-9, find the opposite of each term.

$$2x+3-4x+9=2\sqrt{3x-8}\sqrt{x-1}$$

Combine 2x and -4x to get -2x.

$$-2x+3+9=2\sqrt{3x-8}\sqrt{x-1}$$

Add 3 and 9 to get 12.

$$-2x+12=2\sqrt{3x-8}\sqrt{x-1}$$

Square both sides of the equation.

$$\left(-2x+12\right)^{2}=\left(2\sqrt{3x-8}\sqrt{x-1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x+12\right)^{2}.

$$4x^{2}-48x+144=\left(2\sqrt{3x-8}\sqrt{x-1}\right)^{2}$$

Expand \left(2\sqrt{3x-8}\sqrt{x-1}\right)^{2}.

$$4x^{2}-48x+144=2^{2}\left(\sqrt{3x-8}\right)^{2}\left(\sqrt{x-1}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-48x+144=4\left(\sqrt{3x-8}\right)^{2}\left(\sqrt{x-1}\right)^{2}$$

Calculate \sqrt{3x-8} to the power of 2 and get 3x-8.

$$4x^{2}-48x+144=4\left(3x-8\right)\left(\sqrt{x-1}\right)^{2}$$

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

$$4x^{2}-48x+144=4\left(3x-8\right)\left(x-1\right)$$

Use the distributive property to multiply 4 by 3x-8.

$$4x^{2}-48x+144=\left(12x-32\right)\left(x-1\right)$$

Apply the distributive property by multiplying each term of 12x-32 by each term of x-1.

$$4x^{2}-48x+144=12x^{2}-12x-32x+32$$

Combine -12x and -32x to get -44x.

$$4x^{2}-48x+144=12x^{2}-44x+32$$

Subtract 12x^{2} from both sides.

$$4x^{2}-48x+144-12x^{2}=-44x+32$$

Combine 4x^{2} and -12x^{2} to get -8x^{2}.

$$-8x^{2}-48x+144=-44x+32$$

Add 44x to both sides.

$$-8x^{2}-48x+144+44x=32$$

Combine -48x and 44x to get -4x.

$$-8x^{2}-4x+144=32$$

Subtract 32 from both sides.

$$-8x^{2}-4x+144-32=0$$

Subtract 32 from 144 to get 112.

$$-8x^{2}-4x+112=0$$

Divide both sides by 4.

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

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

$$a+b=-1 ab=-2\times 28=-56$$

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 -56.

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

Calculate the sum for each pair.

$$1-56=-55 2-28=-26 4-14=-10 7-8=-1$$

The solution is the pair that gives sum -1.

$$a=7 b=-8$$

Rewrite -2x^{2}-x+28 as \left(-2x^{2}+7x\right)+\left(-8x+28\right).

$$\left(-2x^{2}+7x\right)+\left(-8x+28\right)$$

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

$$-x\left(2x-7\right)-4\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(-x-4\right)$$

To find equation solutions, solve 2x-7=0 and -x-4=0.

$$x=\frac{7}{2} x=-4$$

Substitute -4 for x in the equation \sqrt{2x+3}-\sqrt{x-1}=\sqrt{3x-8}. The expression \sqrt{2\left(-4\right)+3} is undefined because the radicand cannot be negative.

$$\sqrt{2\left(-4\right)+3}-\sqrt{-4-1}=\sqrt{3\left(-4\right)-8}$$

Substitute \frac{7}{2} for x in the equation \sqrt{2x+3}-\sqrt{x-1}=\sqrt{3x-8}.

$$\sqrt{2\times \frac{7}{2}+3}-\sqrt{\frac{7}{2}-1}=\sqrt{3\times \frac{7}{2}-8}$$

Simplify. The value x=\frac{7}{2} satisfies the equation.

$$\frac{1}{2}\times 10^{\frac{1}{2}}=\frac{1}{2}\times 10^{\frac{1}{2}}$$

Equation \sqrt{2x+3}=\sqrt{x-1}+\sqrt{3x-8} has a unique solution.

$$x=\frac{7}{2}$$