Solve sqrt{4x+1}=2-3x

$$\sqrt { 4 x + 1 } = 2 - 3 x$$

Solve for x

$$x=\frac{8-\sqrt{37}}{9}\approx 0.213026386$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{4x+1} to the power of 2 and get 4x+1.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-3x\right)^{2}.

$$4x+1=4-12x+9x^{2}$$

Subtract 4 from both sides.

$$4x+1-4=-12x+9x^{2}$$

Subtract 4 from 1 to get -3.

$$4x-3=-12x+9x^{2}$$

Add 12x to both sides.

$$4x-3+12x=9x^{2}$$

Combine 4x and 12x to get 16x.

$$16x-3=9x^{2}$$

Subtract 9x^{2} from both sides.

$$16x-3-9x^{2}=0$$

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.

$$-9x^{2}+16x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 16 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-16±\sqrt{256+36\left(-3\right)}}{2\left(-9\right)}$$

Multiply 36 times -3.

$$x=\frac{-16±\sqrt{256-108}}{2\left(-9\right)}$$

Add 256 to -108.

$$x=\frac{-16±\sqrt{148}}{2\left(-9\right)}$$

Take the square root of 148.

$$x=\frac{-16±2\sqrt{37}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-16±2\sqrt{37}}{-18}$$

Now solve the equation x=\frac{-16±2\sqrt{37}}{-18} when ± is plus. Add -16 to 2\sqrt{37}.

$$x=\frac{2\sqrt{37}-16}{-18}$$

Divide -16+2\sqrt{37} by -18.

$$x=\frac{8-\sqrt{37}}{9}$$

Now solve the equation x=\frac{-16±2\sqrt{37}}{-18} when ± is minus. Subtract 2\sqrt{37} from -16.

$$x=\frac{-2\sqrt{37}-16}{-18}$$

Divide -16-2\sqrt{37} by -18.

$$x=\frac{\sqrt{37}+8}{9}$$

The equation is now solved.

$$x=\frac{8-\sqrt{37}}{9} x=\frac{\sqrt{37}+8}{9}$$

Substitute \frac{8-\sqrt{37}}{9} for x in the equation \sqrt{4x+1}=2-3x.

$$\sqrt{4\times \frac{8-\sqrt{37}}{9}+1}=2-3\times \frac{8-\sqrt{37}}{9}$$

Simplify. The value x=\frac{8-\sqrt{37}}{9} satisfies the equation.

$$\frac{1}{3}\times 37^{\frac{1}{2}}-\frac{2}{3}=-\frac{2}{3}+\frac{1}{3}\times 37^{\frac{1}{2}}$$

Substitute \frac{\sqrt{37}+8}{9} for x in the equation \sqrt{4x+1}=2-3x.

$$\sqrt{4\times \frac{\sqrt{37}+8}{9}+1}=2-3\times \frac{\sqrt{37}+8}{9}$$

Simplify. The value x=\frac{\sqrt{37}+8}{9} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{3}\times 37^{\frac{1}{2}}+\frac{2}{3}=-\frac{2}{3}-\frac{1}{3}\times 37^{\frac{1}{2}}$$

Equation \sqrt{4x+1}=2-3x has a unique solution.

$$x=\frac{8-\sqrt{37}}{9}$$