Solve sqrt{2x}+7x=9

$$\sqrt{ 2x } +7x = 9$$

Solve for x

$$x = \frac{64 - \sqrt{127}}{49} \approx 1.076134129$$

Solution Steps

Subtract 7x from both sides of the equation.

$$\sqrt{2x}=9-7x$$

Square both sides of the equation.

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

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

$$2x=\left(9-7x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-7x\right)^{2}.

$$2x=81-126x+49x^{2}$$

Subtract 81 from both sides.

$$2x-81=-126x+49x^{2}$$

Add 126x to both sides.

$$2x-81+126x=49x^{2}$$

Combine 2x and 126x to get 128x.

$$128x-81=49x^{2}$$

Subtract 49x^{2} from both sides.

$$128x-81-49x^{2}=0$$

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.

$$-49x^{2}+128x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -49 for a, 128 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-128±\sqrt{128^{2}-4\left(-49\right)\left(-81\right)}}{2\left(-49\right)}$$

Square 128.

$$x=\frac{-128±\sqrt{16384-4\left(-49\right)\left(-81\right)}}{2\left(-49\right)}$$

Multiply -4 times -49.

$$x=\frac{-128±\sqrt{16384+196\left(-81\right)}}{2\left(-49\right)}$$

Multiply 196 times -81.

$$x=\frac{-128±\sqrt{16384-15876}}{2\left(-49\right)}$$

Add 16384 to -15876.

$$x=\frac{-128±\sqrt{508}}{2\left(-49\right)}$$

Take the square root of 508.

$$x=\frac{-128±2\sqrt{127}}{2\left(-49\right)}$$

Multiply 2 times -49.

$$x=\frac{-128±2\sqrt{127}}{-98}$$

Now solve the equation x=\frac{-128±2\sqrt{127}}{-98} when ± is plus. Add -128 to 2\sqrt{127}.

$$x=\frac{2\sqrt{127}-128}{-98}$$

Divide -128+2\sqrt{127} by -98.

$$x=\frac{64-\sqrt{127}}{49}$$

Now solve the equation x=\frac{-128±2\sqrt{127}}{-98} when ± is minus. Subtract 2\sqrt{127} from -128.

$$x=\frac{-2\sqrt{127}-128}{-98}$$

Divide -128-2\sqrt{127} by -98.

$$x=\frac{\sqrt{127}+64}{49}$$

The equation is now solved.

$$x=\frac{64-\sqrt{127}}{49} x=\frac{\sqrt{127}+64}{49}$$

Substitute \frac{64-\sqrt{127}}{49} for x in the equation \sqrt{2x}+7x=9.

$$\sqrt{2\times \frac{64-\sqrt{127}}{49}}+7\times \frac{64-\sqrt{127}}{49}=9$$

Simplify. The value x=\frac{64-\sqrt{127}}{49} satisfies the equation.

$$9=9$$

Substitute \frac{\sqrt{127}+64}{49} for x in the equation \sqrt{2x}+7x=9.

$$\sqrt{2\times \frac{\sqrt{127}+64}{49}}+7\times \frac{\sqrt{127}+64}{49}=9$$

Simplify. The value x=\frac{\sqrt{127}+64}{49} does not satisfy the equation.

$$\frac{65}{7}+\frac{2}{7}\times 127^{\frac{1}{2}}=9$$

Equation \sqrt{2x}=9-7x has a unique solution.

$$x=\frac{64-\sqrt{127}}{49}$$