Solve sqrt{3x}+4=sqrt{8x+25}

$$\sqrt { 3 x } + 4 = \sqrt { 8 x + 25 }$$

Solve for x

$$x=3$$

$$x = \frac{27}{25} = 1\frac{2}{25} = 1.08$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x}+4\right)^{2}=\left(\sqrt{8x+25}\right)^{2}$$

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

$$\left(\sqrt{3x}\right)^{2}+8\sqrt{3x}+16=\left(\sqrt{8x+25}\right)^{2}$$

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

$$3x+8\sqrt{3x}+16=\left(\sqrt{8x+25}\right)^{2}$$

Calculate \sqrt{8x+25} to the power of 2 and get 8x+25.

$$3x+8\sqrt{3x}+16=8x+25$$

Subtract 3x+16 from both sides of the equation.

$$8\sqrt{3x}=8x+25-\left(3x+16\right)$$

To find the opposite of 3x+16, find the opposite of each term.

$$8\sqrt{3x}=8x+25-3x-16$$

Combine 8x and -3x to get 5x.

$$8\sqrt{3x}=5x+25-16$$

Subtract 16 from 25 to get 9.

$$8\sqrt{3x}=5x+9$$

Square both sides of the equation.

$$\left(8\sqrt{3x}\right)^{2}=\left(5x+9\right)^{2}$$

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

$$8^{2}\left(\sqrt{3x}\right)^{2}=\left(5x+9\right)^{2}$$

Calculate 8 to the power of 2 and get 64.

$$64\left(\sqrt{3x}\right)^{2}=\left(5x+9\right)^{2}$$

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

$$64\times 3x=\left(5x+9\right)^{2}$$

Multiply 64 and 3 to get 192.

$$192x=\left(5x+9\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+9\right)^{2}.

$$192x=25x^{2}+90x+81$$

Subtract 25x^{2} from both sides.

$$192x-25x^{2}=90x+81$$

Subtract 90x from both sides.

$$192x-25x^{2}-90x=81$$

Combine 192x and -90x to get 102x.

$$102x-25x^{2}=81$$

Subtract 81 from both sides.

$$102x-25x^{2}-81=0$$

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.

$$-25x^{2}+102x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 102 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-102±\sqrt{102^{2}-4\left(-25\right)\left(-81\right)}}{2\left(-25\right)}$$

Square 102.

$$x=\frac{-102±\sqrt{10404-4\left(-25\right)\left(-81\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-102±\sqrt{10404+100\left(-81\right)}}{2\left(-25\right)}$$

Multiply 100 times -81.

$$x=\frac{-102±\sqrt{10404-8100}}{2\left(-25\right)}$$

Add 10404 to -8100.

$$x=\frac{-102±\sqrt{2304}}{2\left(-25\right)}$$

Take the square root of 2304.

$$x=\frac{-102±48}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-102±48}{-50}$$

Now solve the equation x=\frac{-102±48}{-50} when ± is plus. Add -102 to 48.

$$x=-\frac{54}{-50}$$

Reduce the fraction \frac{-54}{-50} to lowest terms by extracting and canceling out 2.

$$x=\frac{27}{25}$$

Now solve the equation x=\frac{-102±48}{-50} when ± is minus. Subtract 48 from -102.

$$x=-\frac{150}{-50}$$

Divide -150 by -50.

$$x=3$$

The equation is now solved.

$$x=\frac{27}{25} x=3$$

Substitute \frac{27}{25} for x in the equation \sqrt{3x}+4=\sqrt{8x+25}.

$$\sqrt{3\times \frac{27}{25}}+4=\sqrt{8\times \frac{27}{25}+25}$$

Simplify. The value x=\frac{27}{25} satisfies the equation.

$$\frac{29}{5}=\frac{29}{5}$$

Substitute 3 for x in the equation \sqrt{3x}+4=\sqrt{8x+25}.

$$\sqrt{3\times 3}+4=\sqrt{8\times 3+25}$$

Simplify. The value x=3 satisfies the equation.

$$7=7$$

List all solutions of \sqrt{3x}+4=\sqrt{8x+25}.

$$x=\frac{27}{25} x=3$$