Solve sqrt{7x}-sqrt{5}=sqrt{3x}

$$\sqrt { 7 x } - \sqrt { 5 } = \sqrt { 3 x }$$

Solve for x

$$x = \frac{5 {(\sqrt{21} + 5)}}{8} \approx 5.989109809$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{7x}-\sqrt{5}\right)^{2}=\left(\sqrt{3x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7x}-\sqrt{5}\right)^{2}.

$$\left(\sqrt{7x}\right)^{2}-2\sqrt{7x}\sqrt{5}+\left(\sqrt{5}\right)^{2}=\left(\sqrt{3x}\right)^{2}$$

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

$$7x-2\sqrt{7x}\sqrt{5}+\left(\sqrt{5}\right)^{2}=\left(\sqrt{3x}\right)^{2}$$

The square of \sqrt{5} is 5.

$$7x-2\sqrt{7x}\sqrt{5}+5=\left(\sqrt{3x}\right)^{2}$$

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

$$7x-2\sqrt{7x}\sqrt{5}+5=3x$$

Subtract 7x+5 from both sides of the equation.

$$-2\sqrt{7x}\sqrt{5}=3x-\left(7x+5\right)$$

To find the opposite of 7x+5, find the opposite of each term.

$$-2\sqrt{7x}\sqrt{5}=3x-7x-5$$

Combine 3x and -7x to get -4x.

$$-2\sqrt{7x}\sqrt{5}=-4x-5$$

Square both sides of the equation.

$$\left(-2\sqrt{7x}\sqrt{5}\right)^{2}=\left(-4x-5\right)^{2}$$

Expand \left(-2\sqrt{7x}\sqrt{5}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{7x}\right)^{2}\left(\sqrt{5}\right)^{2}=\left(-4x-5\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{7x}\right)^{2}\left(\sqrt{5}\right)^{2}=\left(-4x-5\right)^{2}$$

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

$$4\times 7x\left(\sqrt{5}\right)^{2}=\left(-4x-5\right)^{2}$$

Multiply 4 and 7 to get 28.

$$28x\left(\sqrt{5}\right)^{2}=\left(-4x-5\right)^{2}$$

The square of \sqrt{5} is 5.

$$28x\times 5=\left(-4x-5\right)^{2}$$

Multiply 28 and 5 to get 140.

$$140x=\left(-4x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4x-5\right)^{2}.

$$140x=16x^{2}+40x+25$$

Subtract 16x^{2} from both sides.

$$140x-16x^{2}=40x+25$$

Subtract 40x from both sides.

$$140x-16x^{2}-40x=25$$

Combine 140x and -40x to get 100x.

$$100x-16x^{2}=25$$

Subtract 25 from both sides.

$$100x-16x^{2}-25=0$$

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.

$$-16x^{2}+100x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 100 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-100±\sqrt{100^{2}-4\left(-16\right)\left(-25\right)}}{2\left(-16\right)}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\left(-16\right)\left(-25\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-100±\sqrt{10000+64\left(-25\right)}}{2\left(-16\right)}$$

Multiply 64 times -25.

$$x=\frac{-100±\sqrt{10000-1600}}{2\left(-16\right)}$$

Add 10000 to -1600.

$$x=\frac{-100±\sqrt{8400}}{2\left(-16\right)}$$

Take the square root of 8400.

$$x=\frac{-100±20\sqrt{21}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-100±20\sqrt{21}}{-32}$$

Now solve the equation x=\frac{-100±20\sqrt{21}}{-32} when ± is plus. Add -100 to 20\sqrt{21}.

$$x=\frac{20\sqrt{21}-100}{-32}$$

Divide -100+20\sqrt{21} by -32.

$$x=\frac{25-5\sqrt{21}}{8}$$

Now solve the equation x=\frac{-100±20\sqrt{21}}{-32} when ± is minus. Subtract 20\sqrt{21} from -100.

$$x=\frac{-20\sqrt{21}-100}{-32}$$

Divide -100-20\sqrt{21} by -32.

$$x=\frac{5\sqrt{21}+25}{8}$$

The equation is now solved.

$$x=\frac{25-5\sqrt{21}}{8} x=\frac{5\sqrt{21}+25}{8}$$

Substitute \frac{25-5\sqrt{21}}{8} for x in the equation \sqrt{7x}-\sqrt{5}=\sqrt{3x}.

$$\sqrt{7\times \frac{25-5\sqrt{21}}{8}}-\sqrt{5}=\sqrt{3\times \frac{25-5\sqrt{21}}{8}}$$

Simplify. The value x=\frac{25-5\sqrt{21}}{8} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{3}{4}\times 5^{\frac{1}{2}}-\frac{1}{4}\times 105^{\frac{1}{2}}=\frac{1}{4}\times 105^{\frac{1}{2}}-\frac{3}{4}\times 5^{\frac{1}{2}}$$

Substitute \frac{5\sqrt{21}+25}{8} for x in the equation \sqrt{7x}-\sqrt{5}=\sqrt{3x}.

$$\sqrt{7\times \frac{5\sqrt{21}+25}{8}}-\sqrt{5}=\sqrt{3\times \frac{5\sqrt{21}+25}{8}}$$

Simplify. The value x=\frac{5\sqrt{21}+25}{8} satisfies the equation.

$$\frac{3}{4}\times 5^{\frac{1}{2}}+\frac{1}{4}\times 105^{\frac{1}{2}}=\frac{1}{4}\times 105^{\frac{1}{2}}+\frac{3}{4}\times 5^{\frac{1}{2}}$$

Equation \sqrt{7x}-\sqrt{5}=\sqrt{3x} has a unique solution.

$$x=\frac{5\sqrt{21}+25}{8}$$