Solve frac{2sqrt{x}+5}{3}=sqrt{x}-1(x)0)

$$\frac { 2 \sqrt { x } + 5 } { 3 } = \sqrt { x } - 1 ( x ) 0 )$$

Solve for x

$$x=25$$

Solution Steps

Subtract 0 from both sides of the equation.

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

Multiply both sides of the equation by 3.

$$2\sqrt{x}+5+3x\times 0=3\sqrt{x}$$

Multiply 3 and 0 to get 0.

$$2\sqrt{x}+5+0x=3\sqrt{x}$$

Anything times zero gives zero.

$$2\sqrt{x}+5+0=3\sqrt{x}$$

Add 5 and 0 to get 5.

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

Square both sides of the equation.

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

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

$$4\left(\sqrt{x}\right)^{2}+20\sqrt{x}+25=\left(3\sqrt{x}\right)^{2}$$

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

$$4x+20\sqrt{x}+25=\left(3\sqrt{x}\right)^{2}$$

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

$$4x+20\sqrt{x}+25=3^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$4x+20\sqrt{x}+25=9\left(\sqrt{x}\right)^{2}$$

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

$$4x+20\sqrt{x}+25=9x$$

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

$$20\sqrt{x}=9x-\left(4x+25\right)$$

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

$$20\sqrt{x}=9x-4x-25$$

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

$$20\sqrt{x}=5x-25$$

Square both sides of the equation.

$$\left(20\sqrt{x}\right)^{2}=\left(5x-25\right)^{2}$$

Expand \left(20\sqrt{x}\right)^{2}.

$$20^{2}\left(\sqrt{x}\right)^{2}=\left(5x-25\right)^{2}$$

Calculate 20 to the power of 2 and get 400.

$$400\left(\sqrt{x}\right)^{2}=\left(5x-25\right)^{2}$$

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

$$400x=\left(5x-25\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-25\right)^{2}.

$$400x=25x^{2}-250x+625$$

Subtract 25x^{2} from both sides.

$$400x-25x^{2}=-250x+625$$

Add 250x to both sides.

$$400x-25x^{2}+250x=625$$

Combine 400x and 250x to get 650x.

$$650x-25x^{2}=625$$

Subtract 625 from both sides.

$$650x-25x^{2}-625=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-25x^{2}+650x-625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 650 for b, and -625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-650±\sqrt{650^{2}-4\left(-25\right)\left(-625\right)}}{2\left(-25\right)}$$

Square 650.

$$x=\frac{-650±\sqrt{422500-4\left(-25\right)\left(-625\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-650±\sqrt{422500+100\left(-625\right)}}{2\left(-25\right)}$$

Multiply 100 times -625.

$$x=\frac{-650±\sqrt{422500-62500}}{2\left(-25\right)}$$

Add 422500 to -62500.

$$x=\frac{-650±\sqrt{360000}}{2\left(-25\right)}$$

Take the square root of 360000.

$$x=\frac{-650±600}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-650±600}{-50}$$

Now solve the equation x=\frac{-650±600}{-50} when ± is plus. Add -650 to 600.

$$x=-\frac{50}{-50}$$

Divide -50 by -50.

$$x=1$$

Now solve the equation x=\frac{-650±600}{-50} when ± is minus. Subtract 600 from -650.

$$x=-\frac{1250}{-50}$$

Divide -1250 by -50.

$$x=25$$

The equation is now solved.

$$x=1 x=25$$

Substitute 1 for x in the equation \frac{2\sqrt{x}+5}{3}=\sqrt{x}-1x\times 0.

$$\frac{2\sqrt{1}+5}{3}=\sqrt{1}-1\times 1\times 0$$

Simplify. The value x=1 does not satisfy the equation.

$$\frac{7}{3}=1$$

Substitute 25 for x in the equation \frac{2\sqrt{x}+5}{3}=\sqrt{x}-1x\times 0.

$$\frac{2\sqrt{25}+5}{3}=\sqrt{25}-1\times 25\times 0$$

Simplify. The value x=25 satisfies the equation.

$$5=5$$

Equation 2\sqrt{x}+5=3\sqrt{x} has a unique solution.

$$x=25$$