Solve sqrt{x}+3x+5=3

$$\sqrt{ x } +3x+5=3$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{23}i-11}{18}\approx -0.611111111-0.266435085i$$

Solution Steps

Subtract 3x+5 from both sides of the equation.

$$\sqrt{x}=3-\left(3x+5\right)$$

To find the opposite of 3x+5, find the opposite of each term.

$$\sqrt{x}=3-3x-5$$

Subtract 5 from 3 to get -2.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 4 from both sides.

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

Subtract 12x from both sides.

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

Combine x and -12x to get -11x.

$$-11x-4=9x^{2}$$

Subtract 9x^{2} from both sides.

$$-11x-4-9x^{2}=0$$

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.

$$-9x^{2}-11x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -11 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-9\right)\left(-4\right)}}{2\left(-9\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-9\right)\left(-4\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-11\right)±\sqrt{121+36\left(-4\right)}}{2\left(-9\right)}$$

Multiply 36 times -4.

$$x=\frac{-\left(-11\right)±\sqrt{121-144}}{2\left(-9\right)}$$

Add 121 to -144.

$$x=\frac{-\left(-11\right)±\sqrt{-23}}{2\left(-9\right)}$$

Take the square root of -23.

$$x=\frac{-\left(-11\right)±\sqrt{23}i}{2\left(-9\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{23}i}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{11±\sqrt{23}i}{-18}$$

Now solve the equation x=\frac{11±\sqrt{23}i}{-18} when ± is plus. Add 11 to i\sqrt{23}.

$$x=\frac{11+\sqrt{23}i}{-18}$$

Divide 11+i\sqrt{23} by -18.

$$x=\frac{-\sqrt{23}i-11}{18}$$

Now solve the equation x=\frac{11±\sqrt{23}i}{-18} when ± is minus. Subtract i\sqrt{23} from 11.

$$x=\frac{-\sqrt{23}i+11}{-18}$$

Divide 11-i\sqrt{23} by -18.

$$x=\frac{-11+\sqrt{23}i}{18}$$

The equation is now solved.

$$x=\frac{-\sqrt{23}i-11}{18} x=\frac{-11+\sqrt{23}i}{18}$$

Substitute \frac{-\sqrt{23}i-11}{18} for x in the equation \sqrt{x}+3x+5=3.

$$\sqrt{\frac{-\sqrt{23}i-11}{18}}+3\times \frac{-\sqrt{23}i-11}{18}+5=3$$

Simplify. The value x=\frac{-\sqrt{23}i-11}{18} satisfies the equation.

$$3=3$$

Substitute \frac{-11+\sqrt{23}i}{18} for x in the equation \sqrt{x}+3x+5=3.

$$\sqrt{\frac{-11+\sqrt{23}i}{18}}+3\times \frac{-11+\sqrt{23}i}{18}+5=3$$

Simplify. The value x=\frac{-11+\sqrt{23}i}{18} does not satisfy the equation.

$$\frac{10}{3}+\frac{1}{3}i\times 23^{\frac{1}{2}}=3$$

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

$$x=\frac{-\sqrt{23}i-11}{18}$$