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

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

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{7}i}{8}\approx -0.375+0.330718914i$$

Solution Steps

Subtract -2x+3 from both sides of the equation.

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

To find the opposite of -2x+3, find the opposite of each term.

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

The opposite of -2x is 2x.

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

Subtract 3 from 4 to get 1.

$$\sqrt{x}=1+2x$$

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+2x\right)^{2}.

$$x=1+4x+4x^{2}$$

Subtract 1 from both sides.

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

Subtract 4x from both sides.

$$x-1-4x=4x^{2}$$

Combine x and -4x to get -3x.

$$-3x-1=4x^{2}$$

Subtract 4x^{2} from both sides.

$$-3x-1-4x^{2}=0$$

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.

$$-4x^{2}-3x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16\left(-1\right)}}{2\left(-4\right)}$$

Multiply 16 times -1.

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

Add 9 to -16.

$$x=\frac{-\left(-3\right)±\sqrt{-7}}{2\left(-4\right)}$$

Take the square root of -7.

$$x=\frac{-\left(-3\right)±\sqrt{7}i}{2\left(-4\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{7}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{3±\sqrt{7}i}{-8}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{-8} when ± is plus. Add 3 to i\sqrt{7}.

$$x=\frac{3+\sqrt{7}i}{-8}$$

Divide 3+i\sqrt{7} by -8.

$$x=\frac{-\sqrt{7}i-3}{8}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{-8} when ± is minus. Subtract i\sqrt{7} from 3.

$$x=\frac{-\sqrt{7}i+3}{-8}$$

Divide 3-i\sqrt{7} by -8.

$$x=\frac{-3+\sqrt{7}i}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}i-3}{8} x=\frac{-3+\sqrt{7}i}{8}$$

Substitute \frac{-\sqrt{7}i-3}{8} for x in the equation \sqrt{x}-2x+3=4.

$$\sqrt{\frac{-\sqrt{7}i-3}{8}}-2\times \frac{-\sqrt{7}i-3}{8}+3=4$$

Simplify. The value x=\frac{-\sqrt{7}i-3}{8} does not satisfy the equation.

$$\frac{7}{2}+\frac{1}{2}i\times 7^{\frac{1}{2}}=4$$

Substitute \frac{-3+\sqrt{7}i}{8} for x in the equation \sqrt{x}-2x+3=4.

$$\sqrt{\frac{-3+\sqrt{7}i}{8}}-2\times \frac{-3+\sqrt{7}i}{8}+3=4$$

Simplify. The value x=\frac{-3+\sqrt{7}i}{8} satisfies the equation.

$$4=4$$

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

$$x=\frac{-3+\sqrt{7}i}{8}$$