Solve sqrt{2x^2-5x+12+2x^2}=5x

$$\sqrt { 2 x ^ { 2 } - 5 x + 12 + 2 x ^ { 2 } } = 5 x$$

Solve for x

$$x=\frac{\sqrt{1033}-5}{42}\approx 0.646198032$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x^{2}-5x+12+2x^{2}}\right)^{2}=\left(5x\right)^{2}$$

Combine 2x^{2} and 2x^{2} to get 4x^{2}.

$$\left(\sqrt{4x^{2}-5x+12}\right)^{2}=\left(5x\right)^{2}$$

Calculate \sqrt{4x^{2}-5x+12} to the power of 2 and get 4x^{2}-5x+12.

$$4x^{2}-5x+12=\left(5x\right)^{2}$$

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

$$4x^{2}-5x+12=5^{2}x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$4x^{2}-5x+12=25x^{2}$$

Subtract 25x^{2} from both sides.

$$4x^{2}-5x+12-25x^{2}=0$$

Combine 4x^{2} and -25x^{2} to get -21x^{2}.

$$-21x^{2}-5x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -21 for a, -5 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-21\right)\times 12}}{2\left(-21\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-21\right)\times 12}}{2\left(-21\right)}$$

Multiply -4 times -21.

$$x=\frac{-\left(-5\right)±\sqrt{25+84\times 12}}{2\left(-21\right)}$$

Multiply 84 times 12.

$$x=\frac{-\left(-5\right)±\sqrt{25+1008}}{2\left(-21\right)}$$

Add 25 to 1008.

$$x=\frac{-\left(-5\right)±\sqrt{1033}}{2\left(-21\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{1033}}{2\left(-21\right)}$$

Multiply 2 times -21.

$$x=\frac{5±\sqrt{1033}}{-42}$$

Now solve the equation x=\frac{5±\sqrt{1033}}{-42} when ± is plus. Add 5 to \sqrt{1033}.

$$x=\frac{\sqrt{1033}+5}{-42}$$

Divide 5+\sqrt{1033} by -42.

$$x=\frac{-\sqrt{1033}-5}{42}$$

Now solve the equation x=\frac{5±\sqrt{1033}}{-42} when ± is minus. Subtract \sqrt{1033} from 5.

$$x=\frac{5-\sqrt{1033}}{-42}$$

Divide 5-\sqrt{1033} by -42.

$$x=\frac{\sqrt{1033}-5}{42}$$

The equation is now solved.

$$x=\frac{-\sqrt{1033}-5}{42} x=\frac{\sqrt{1033}-5}{42}$$

Substitute \frac{-\sqrt{1033}-5}{42} for x in the equation \sqrt{2x^{2}-5x+12+2x^{2}}=5x.

$$\sqrt{2\times \left(\frac{-\sqrt{1033}-5}{42}\right)^{2}-5\times \frac{-\sqrt{1033}-5}{42}+12+2\times \left(\frac{-\sqrt{1033}-5}{42}\right)^{2}}=5\times \frac{-\sqrt{1033}-5}{42}$$

Simplify. The value x=\frac{-\sqrt{1033}-5}{42} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{25}{42}+\frac{5}{42}\times 1033^{\frac{1}{2}}=-\frac{5}{42}\times 1033^{\frac{1}{2}}-\frac{25}{42}$$

Substitute \frac{\sqrt{1033}-5}{42} for x in the equation \sqrt{2x^{2}-5x+12+2x^{2}}=5x.

$$\sqrt{2\times \left(\frac{\sqrt{1033}-5}{42}\right)^{2}-5\times \frac{\sqrt{1033}-5}{42}+12+2\times \left(\frac{\sqrt{1033}-5}{42}\right)^{2}}=5\times \frac{\sqrt{1033}-5}{42}$$

Simplify. The value x=\frac{\sqrt{1033}-5}{42} satisfies the equation.

$$-\left(\frac{25}{42}-\frac{5}{42}\times 1033^{\frac{1}{2}}\right)=\frac{5}{42}\times 1033^{\frac{1}{2}}-\frac{25}{42}$$

Equation \sqrt{2x^{2}+2x^{2}-5x+12}=5x has a unique solution.

$$x=\frac{\sqrt{1033}-5}{42}$$