Solve frac{3sqrt{x}-5}{2}=sqrt{x}-2

$$\frac { 3 \sqrt { x } - 5 } { 2 } = \sqrt { x } - 2$$

Solve for x

$$x=1$$

Solution Steps

Subtract -2 from both sides of the equation.

$$\frac{3\sqrt{x}-5}{2}+2=\sqrt{x}$$

Multiply both sides of the equation by 2.

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

Add -5 and 4 to get -1.

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

Square both sides of the equation.

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

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

$$9\left(\sqrt{x}\right)^{2}-6\sqrt{x}+1=\left(2\sqrt{x}\right)^{2}$$

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

$$9x-6\sqrt{x}+1=\left(2\sqrt{x}\right)^{2}$$

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

$$9x-6\sqrt{x}+1=2^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x-6\sqrt{x}+1=4\left(\sqrt{x}\right)^{2}$$

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

$$9x-6\sqrt{x}+1=4x$$

Subtract 9x+1 from both sides of the equation.

$$-6\sqrt{x}=4x-\left(9x+1\right)$$

To find the opposite of 9x+1, find the opposite of each term.

$$-6\sqrt{x}=4x-9x-1$$

Combine 4x and -9x to get -5x.

$$-6\sqrt{x}=-5x-1$$

Square both sides of the equation.

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

Expand \left(-6\sqrt{x}\right)^{2}.

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

Calculate -6 to the power of 2 and get 36.

$$36\left(\sqrt{x}\right)^{2}=\left(-5x-1\right)^{2}$$

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

$$36x=\left(-5x-1\right)^{2}$$

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

$$36x=25x^{2}+10x+1$$

Subtract 25x^{2} from both sides.

$$36x-25x^{2}=10x+1$$

Subtract 10x from both sides.

$$36x-25x^{2}-10x=1$$

Combine 36x and -10x to get 26x.

$$26x-25x^{2}=1$$

Subtract 1 from both sides.

$$26x-25x^{2}-1=0$$

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

$$-25x^{2}+26x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -25x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=26 ab=-25\left(-1\right)=25$$

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 25.

$$1,25 5,5$$

Calculate the sum for each pair.

$$1+25=26 5+5=10$$

The solution is the pair that gives sum 26.

$$a=25 b=1$$

Rewrite -25x^{2}+26x-1 as \left(-25x^{2}+25x\right)+\left(x-1\right).

$$\left(-25x^{2}+25x\right)+\left(x-1\right)$$

Factor out 25x in the first and -1 in the second group.

$$25x\left(-x+1\right)-\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(25x-1\right)$$

To find equation solutions, solve -x+1=0 and 25x-1=0.

$$x=1 x=\frac{1}{25}$$

Substitute 1 for x in the equation \frac{3\sqrt{x}-5}{2}=\sqrt{x}-2.

$$\frac{3\sqrt{1}-5}{2}=\sqrt{1}-2$$

Simplify. The value x=1 satisfies the equation.

$$-1=-1$$

Substitute \frac{1}{25} for x in the equation \frac{3\sqrt{x}-5}{2}=\sqrt{x}-2.

$$\frac{3\sqrt{\frac{1}{25}}-5}{2}=\sqrt{\frac{1}{25}}-2$$

Simplify. The value x=\frac{1}{25} does not satisfy the equation.

$$-\frac{11}{5}=-\frac{9}{5}$$

Substitute 1 for x in the equation \frac{3\sqrt{x}-5}{2}=\sqrt{x}-2.

$$\frac{3\sqrt{1}-5}{2}=\sqrt{1}-2$$

Simplify. The value x=1 satisfies the equation.

$$-1=-1$$

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

$$x=1$$