Solve 4x+sqrt{x}-6=0

$$4x+ \sqrt{ x } -6=0$$

Solve for x

$$x = \frac{49 - \sqrt{97}}{32} \approx 1.223473194$$

Solution Steps

Subtract 4x-6 from both sides of the equation.

$$\sqrt{x}=-\left(4x-6\right)$$

To find the opposite of 4x-6, find the opposite of each term.

$$\sqrt{x}=-4x-\left(-6\right)$$

The opposite of -6 is 6.

$$\sqrt{x}=-4x+6$$

Square both sides of the equation.

$$\left(\sqrt{x}\right)^{2}=\left(-4x+6\right)^{2}$$

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

$$x=\left(-4x+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4x+6\right)^{2}.

$$x=16x^{2}-48x+36$$

Subtract 16x^{2} from both sides.

$$x-16x^{2}=-48x+36$$

Add 48x to both sides.

$$x-16x^{2}+48x=36$$

Combine x and 48x to get 49x.

$$49x-16x^{2}=36$$

Subtract 36 from both sides.

$$49x-16x^{2}-36=0$$

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.

$$-16x^{2}+49x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 49 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-49±\sqrt{49^{2}-4\left(-16\right)\left(-36\right)}}{2\left(-16\right)}$$

Square 49.

$$x=\frac{-49±\sqrt{2401-4\left(-16\right)\left(-36\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-49±\sqrt{2401+64\left(-36\right)}}{2\left(-16\right)}$$

Multiply 64 times -36.

$$x=\frac{-49±\sqrt{2401-2304}}{2\left(-16\right)}$$

Add 2401 to -2304.

$$x=\frac{-49±\sqrt{97}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-49±\sqrt{97}}{-32}$$

Now solve the equation x=\frac{-49±\sqrt{97}}{-32} when ± is plus. Add -49 to \sqrt{97}.

$$x=\frac{\sqrt{97}-49}{-32}$$

Divide -49+\sqrt{97} by -32.

$$x=\frac{49-\sqrt{97}}{32}$$

Now solve the equation x=\frac{-49±\sqrt{97}}{-32} when ± is minus. Subtract \sqrt{97} from -49.

$$x=\frac{-\sqrt{97}-49}{-32}$$

Divide -49-\sqrt{97} by -32.

$$x=\frac{\sqrt{97}+49}{32}$$

The equation is now solved.

$$x=\frac{49-\sqrt{97}}{32} x=\frac{\sqrt{97}+49}{32}$$

Substitute \frac{49-\sqrt{97}}{32} for x in the equation 4x+\sqrt{x}-6=0.

$$4\times \frac{49-\sqrt{97}}{32}+\sqrt{\frac{49-\sqrt{97}}{32}}-6=0$$

Simplify. The value x=\frac{49-\sqrt{97}}{32} satisfies the equation.

$$0=0$$

Substitute \frac{\sqrt{97}+49}{32} for x in the equation 4x+\sqrt{x}-6=0.

$$4\times \frac{\sqrt{97}+49}{32}+\sqrt{\frac{\sqrt{97}+49}{32}}-6=0$$

Simplify. The value x=\frac{\sqrt{97}+49}{32} does not satisfy the equation.

$$\frac{1}{4}\times 97^{\frac{1}{2}}+\frac{1}{4}=0$$

Equation \sqrt{x}=6-4x has a unique solution.

$$x=\frac{49-\sqrt{97}}{32}$$