Solve sqrt{7x}+2=3sqrt{3x-2}

$$\sqrt { 7 x } + 2 = 3 \sqrt { 3 x - 2 }$$

Solve for x

$$x=\frac{3\sqrt{91}}{50}+\frac{31}{25}\approx 1.812363521$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{7x}+2\right)^{2}=\left(3\sqrt{3x-2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7x}+2\right)^{2}.

$$\left(\sqrt{7x}\right)^{2}+4\sqrt{7x}+4=\left(3\sqrt{3x-2}\right)^{2}$$

Calculate \sqrt{7x} to the power of 2 and get 7x.

$$7x+4\sqrt{7x}+4=\left(3\sqrt{3x-2}\right)^{2}$$

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

$$7x+4\sqrt{7x}+4=3^{2}\left(\sqrt{3x-2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$7x+4\sqrt{7x}+4=9\left(\sqrt{3x-2}\right)^{2}$$

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

$$7x+4\sqrt{7x}+4=9\left(3x-2\right)$$

Use the distributive property to multiply 9 by 3x-2.

$$7x+4\sqrt{7x}+4=27x-18$$

Subtract 7x+4 from both sides of the equation.

$$4\sqrt{7x}=27x-18-\left(7x+4\right)$$

To find the opposite of 7x+4, find the opposite of each term.

$$4\sqrt{7x}=27x-18-7x-4$$

Combine 27x and -7x to get 20x.

$$4\sqrt{7x}=20x-18-4$$

Subtract 4 from -18 to get -22.

$$4\sqrt{7x}=20x-22$$

Square both sides of the equation.

$$\left(4\sqrt{7x}\right)^{2}=\left(20x-22\right)^{2}$$

Expand \left(4\sqrt{7x}\right)^{2}.

$$4^{2}\left(\sqrt{7x}\right)^{2}=\left(20x-22\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{7x}\right)^{2}=\left(20x-22\right)^{2}$$

Calculate \sqrt{7x} to the power of 2 and get 7x.

$$16\times 7x=\left(20x-22\right)^{2}$$

Multiply 16 and 7 to get 112.

$$112x=\left(20x-22\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(20x-22\right)^{2}.

$$112x=400x^{2}-880x+484$$

Subtract 400x^{2} from both sides.

$$112x-400x^{2}=-880x+484$$

Add 880x to both sides.

$$112x-400x^{2}+880x=484$$

Combine 112x and 880x to get 992x.

$$992x-400x^{2}=484$$

Subtract 484 from both sides.

$$992x-400x^{2}-484=0$$

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.

$$-400x^{2}+992x-484=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -400 for a, 992 for b, and -484 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-992±\sqrt{992^{2}-4\left(-400\right)\left(-484\right)}}{2\left(-400\right)}$$

Square 992.

$$x=\frac{-992±\sqrt{984064-4\left(-400\right)\left(-484\right)}}{2\left(-400\right)}$$

Multiply -4 times -400.

$$x=\frac{-992±\sqrt{984064+1600\left(-484\right)}}{2\left(-400\right)}$$

Multiply 1600 times -484.

$$x=\frac{-992±\sqrt{984064-774400}}{2\left(-400\right)}$$

Add 984064 to -774400.

$$x=\frac{-992±\sqrt{209664}}{2\left(-400\right)}$$

Take the square root of 209664.

$$x=\frac{-992±48\sqrt{91}}{2\left(-400\right)}$$

Multiply 2 times -400.

$$x=\frac{-992±48\sqrt{91}}{-800}$$

Now solve the equation x=\frac{-992±48\sqrt{91}}{-800} when ± is plus. Add -992 to 48\sqrt{91}.

$$x=\frac{48\sqrt{91}-992}{-800}$$

Divide -992+48\sqrt{91} by -800.

$$x=-\frac{3\sqrt{91}}{50}+\frac{31}{25}$$

Now solve the equation x=\frac{-992±48\sqrt{91}}{-800} when ± is minus. Subtract 48\sqrt{91} from -992.

$$x=\frac{-48\sqrt{91}-992}{-800}$$

Divide -992-48\sqrt{91} by -800.

$$x=\frac{3\sqrt{91}}{50}+\frac{31}{25}$$

The equation is now solved.

$$x=-\frac{3\sqrt{91}}{50}+\frac{31}{25} x=\frac{3\sqrt{91}}{50}+\frac{31}{25}$$

Substitute -\frac{3\sqrt{91}}{50}+\frac{31}{25} for x in the equation \sqrt{7x}+2=3\sqrt{3x-2}.

$$\sqrt{7\left(-\frac{3\sqrt{91}}{50}+\frac{31}{25}\right)}+2=3\sqrt{3\left(-\frac{3\sqrt{91}}{50}+\frac{31}{25}\right)-2}$$

Simplify. The value x=-\frac{3\sqrt{91}}{50}+\frac{31}{25} does not satisfy the equation.

$$\frac{13}{10}+\frac{3}{10}\times 91^{\frac{1}{2}}=-\frac{27}{10}+\frac{3}{10}\times 91^{\frac{1}{2}}$$

Substitute \frac{3\sqrt{91}}{50}+\frac{31}{25} for x in the equation \sqrt{7x}+2=3\sqrt{3x-2}.

$$\sqrt{7\left(\frac{3\sqrt{91}}{50}+\frac{31}{25}\right)}+2=3\sqrt{3\left(\frac{3\sqrt{91}}{50}+\frac{31}{25}\right)-2}$$

Simplify. The value x=\frac{3\sqrt{91}}{50}+\frac{31}{25} satisfies the equation.

$$\frac{27}{10}+\frac{3}{10}\times 91^{\frac{1}{2}}=\frac{27}{10}+\frac{3}{10}\times 91^{\frac{1}{2}}$$

Equation \sqrt{7x}+2=3\sqrt{3x-2} has a unique solution.

$$x=\frac{3\sqrt{91}}{50}+\frac{31}{25}$$