Solve frac{sqrt{2x^2+5}}{4x+1}=5/3

$$\frac { \sqrt { 2 x ^ { 2 } + 5 } } { 4 x + 1 } = \frac { 5 } { 3 }$$

Solve for x

$$x=\frac{21\sqrt{10}-50}{191}\approx 0.085904874$$

Solution Steps

Variable x cannot be equal to -\frac{1}{4} since division by zero is not defined. Multiply both sides of the equation by 3\left(4x+1\right), the least common multiple of 4x+1,3.

$$3\sqrt{2x^{2}+5}=5\left(4x+1\right)$$

Use the distributive property to multiply 5 by 4x+1.

$$3\sqrt{2x^{2}+5}=20x+5$$

Subtract 20x from both sides.

$$3\sqrt{2x^{2}+5}-20x=5$$

Subtract -20x from both sides of the equation.

$$3\sqrt{2x^{2}+5}=5+20x$$

Square both sides of the equation.

$$\left(3\sqrt{2x^{2}+5}\right)^{2}=\left(20x+5\right)^{2}$$

Expand \left(3\sqrt{2x^{2}+5}\right)^{2}.

$$3^{2}\left(\sqrt{2x^{2}+5}\right)^{2}=\left(20x+5\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{2x^{2}+5}\right)^{2}=\left(20x+5\right)^{2}$$

Calculate \sqrt{2x^{2}+5} to the power of 2 and get 2x^{2}+5.

$$9\left(2x^{2}+5\right)=\left(20x+5\right)^{2}$$

Use the distributive property to multiply 9 by 2x^{2}+5.

$$18x^{2}+45=\left(20x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(20x+5\right)^{2}.

$$18x^{2}+45=400x^{2}+200x+25$$

Subtract 400x^{2} from both sides.

$$18x^{2}+45-400x^{2}=200x+25$$

Combine 18x^{2} and -400x^{2} to get -382x^{2}.

$$-382x^{2}+45=200x+25$$

Subtract 200x from both sides.

$$-382x^{2}+45-200x=25$$

Subtract 25 from both sides.

$$-382x^{2}+45-200x-25=0$$

Subtract 25 from 45 to get 20.

$$-382x^{2}+20-200x=0$$

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.

$$-382x^{2}-200x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -382 for a, -200 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-200\right)±\sqrt{\left(-200\right)^{2}-4\left(-382\right)\times 20}}{2\left(-382\right)}$$

Square -200.

$$x=\frac{-\left(-200\right)±\sqrt{40000-4\left(-382\right)\times 20}}{2\left(-382\right)}$$

Multiply -4 times -382.

$$x=\frac{-\left(-200\right)±\sqrt{40000+1528\times 20}}{2\left(-382\right)}$$

Multiply 1528 times 20.

$$x=\frac{-\left(-200\right)±\sqrt{40000+30560}}{2\left(-382\right)}$$

Add 40000 to 30560.

$$x=\frac{-\left(-200\right)±\sqrt{70560}}{2\left(-382\right)}$$

Take the square root of 70560.

$$x=\frac{-\left(-200\right)±84\sqrt{10}}{2\left(-382\right)}$$

The opposite of -200 is 200.

$$x=\frac{200±84\sqrt{10}}{2\left(-382\right)}$$

Multiply 2 times -382.

$$x=\frac{200±84\sqrt{10}}{-764}$$

Now solve the equation x=\frac{200±84\sqrt{10}}{-764} when ± is plus. Add 200 to 84\sqrt{10}.

$$x=\frac{84\sqrt{10}+200}{-764}$$

Divide 200+84\sqrt{10} by -764.

$$x=\frac{-21\sqrt{10}-50}{191}$$

Now solve the equation x=\frac{200±84\sqrt{10}}{-764} when ± is minus. Subtract 84\sqrt{10} from 200.

$$x=\frac{200-84\sqrt{10}}{-764}$$

Divide 200-84\sqrt{10} by -764.

$$x=\frac{21\sqrt{10}-50}{191}$$

The equation is now solved.

$$x=\frac{-21\sqrt{10}-50}{191} x=\frac{21\sqrt{10}-50}{191}$$

Substitute \frac{-21\sqrt{10}-50}{191} for x in the equation \frac{\sqrt{2x^{2}+5}}{4x+1}=\frac{5}{3}.

$$\frac{\sqrt{2\times \left(\frac{-21\sqrt{10}-50}{191}\right)^{2}+5}}{4\times \frac{-21\sqrt{10}-50}{191}+1}=\frac{5}{3}$$

Simplify. The value x=\frac{-21\sqrt{10}-50}{191} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{5}{3}=\frac{5}{3}$$

Substitute \frac{21\sqrt{10}-50}{191} for x in the equation \frac{\sqrt{2x^{2}+5}}{4x+1}=\frac{5}{3}.

$$\frac{\sqrt{2\times \left(\frac{21\sqrt{10}-50}{191}\right)^{2}+5}}{4\times \frac{21\sqrt{10}-50}{191}+1}=\frac{5}{3}$$

Simplify. The value x=\frac{21\sqrt{10}-50}{191} satisfies the equation.

$$\frac{5}{3}=\frac{5}{3}$$

Equation 3\sqrt{2x^{2}+5}=20x+5 has a unique solution.

$$x=\frac{21\sqrt{10}-50}{191}$$