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

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

Solve for x (complex solution)

$$x=\frac{-29+\sqrt{167}i}{18}\approx -1.611111111+0.717935999i$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x-3}\right)^{2}=\left(3x+5\right)^{2}$$

Calculate \sqrt{x-3} to the power of 2 and get x-3.

$$x-3=\left(3x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$x-3=9x^{2}+30x+25$$

Subtract 9x^{2} from both sides.

$$x-3-9x^{2}=30x+25$$

Subtract 30x from both sides.

$$x-3-9x^{2}-30x=25$$

Combine x and -30x to get -29x.

$$-29x-3-9x^{2}=25$$

Subtract 25 from both sides.

$$-29x-3-9x^{2}-25=0$$

Subtract 25 from -3 to get -28.

$$-29x-28-9x^{2}=0$$

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.

$$-9x^{2}-29x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -29 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\left(-9\right)\left(-28\right)}}{2\left(-9\right)}$$

Square -29.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\left(-9\right)\left(-28\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-29\right)±\sqrt{841+36\left(-28\right)}}{2\left(-9\right)}$$

Multiply 36 times -28.

$$x=\frac{-\left(-29\right)±\sqrt{841-1008}}{2\left(-9\right)}$$

Add 841 to -1008.

$$x=\frac{-\left(-29\right)±\sqrt{-167}}{2\left(-9\right)}$$

Take the square root of -167.

$$x=\frac{-\left(-29\right)±\sqrt{167}i}{2\left(-9\right)}$$

The opposite of -29 is 29.

$$x=\frac{29±\sqrt{167}i}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{29±\sqrt{167}i}{-18}$$

Now solve the equation x=\frac{29±\sqrt{167}i}{-18} when ± is plus. Add 29 to i\sqrt{167}.

$$x=\frac{29+\sqrt{167}i}{-18}$$

Divide 29+i\sqrt{167} by -18.

$$x=\frac{-\sqrt{167}i-29}{18}$$

Now solve the equation x=\frac{29±\sqrt{167}i}{-18} when ± is minus. Subtract i\sqrt{167} from 29.

$$x=\frac{-\sqrt{167}i+29}{-18}$$

Divide 29-i\sqrt{167} by -18.

$$x=\frac{-29+\sqrt{167}i}{18}$$

The equation is now solved.

$$x=\frac{-\sqrt{167}i-29}{18} x=\frac{-29+\sqrt{167}i}{18}$$

Substitute \frac{-\sqrt{167}i-29}{18} for x in the equation \sqrt{x-3}=3x+5.

$$\sqrt{\frac{-\sqrt{167}i-29}{18}-3}=3\times \frac{-\sqrt{167}i-29}{18}+5$$

Simplify. The value x=\frac{-\sqrt{167}i-29}{18} does not satisfy the equation.

$$-\left(\frac{1}{6}-\frac{1}{6}i\times 167^{\frac{1}{2}}\right)=-\frac{1}{6}i\times 167^{\frac{1}{2}}+\frac{1}{6}$$

Substitute \frac{-29+\sqrt{167}i}{18} for x in the equation \sqrt{x-3}=3x+5.

$$\sqrt{\frac{-29+\sqrt{167}i}{18}-3}=3\times \frac{-29+\sqrt{167}i}{18}+5$$

Simplify. The value x=\frac{-29+\sqrt{167}i}{18} satisfies the equation.

$$\frac{1}{6}+\frac{1}{6}i\times 167^{\frac{1}{2}}=\frac{1}{6}+\frac{1}{6}i\times 167^{\frac{1}{2}}$$

Equation \sqrt{x-3}=3x+5 has a unique solution.

$$x=\frac{-29+\sqrt{167}i}{18}$$