Solve sqrt{2x}+x-1=5

$$\sqrt { 2 x } + x - 1 = 5$$

Solve for x

$$x=7-\sqrt{13}\approx 3.394448725$$

Solution Steps

Subtract x-1 from both sides of the equation.

$$\sqrt{2x}=5-\left(x-1\right)$$

To find the opposite of x-1, find the opposite of each term.

$$\sqrt{2x}=5-x-\left(-1\right)$$

The opposite of -1 is 1.

$$\sqrt{2x}=5-x+1$$

Add 5 and 1 to get 6.

$$\sqrt{2x}=6-x$$

Square both sides of the equation.

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

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

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

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

$$2x=36-12x+x^{2}$$

Subtract 36 from both sides.

$$2x-36=-12x+x^{2}$$

Add 12x to both sides.

$$2x-36+12x=x^{2}$$

Combine 2x and 12x to get 14x.

$$14x-36=x^{2}$$

Subtract x^{2} from both sides.

$$14x-36-x^{2}=0$$

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.

$$-x^{2}+14x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 14 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-14±\sqrt{196+4\left(-36\right)}}{2\left(-1\right)}$$

Multiply 4 times -36.

$$x=\frac{-14±\sqrt{196-144}}{2\left(-1\right)}$$

Add 196 to -144.

$$x=\frac{-14±\sqrt{52}}{2\left(-1\right)}$$

Take the square root of 52.

$$x=\frac{-14±2\sqrt{13}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-14±2\sqrt{13}}{-2}$$

Now solve the equation x=\frac{-14±2\sqrt{13}}{-2} when ± is plus. Add -14 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-14}{-2}$$

Divide -14+2\sqrt{13} by -2.

$$x=7-\sqrt{13}$$

Now solve the equation x=\frac{-14±2\sqrt{13}}{-2} when ± is minus. Subtract 2\sqrt{13} from -14.

$$x=\frac{-2\sqrt{13}-14}{-2}$$

Divide -14-2\sqrt{13} by -2.

$$x=\sqrt{13}+7$$

The equation is now solved.

$$x=7-\sqrt{13} x=\sqrt{13}+7$$

Substitute 7-\sqrt{13} for x in the equation \sqrt{2x}+x-1=5.

$$\sqrt{2\left(7-\sqrt{13}\right)}+7-\sqrt{13}-1=5$$

Simplify. The value x=7-\sqrt{13} satisfies the equation.

$$5=5$$

Substitute \sqrt{13}+7 for x in the equation \sqrt{2x}+x-1=5.

$$\sqrt{2\left(\sqrt{13}+7\right)}+\sqrt{13}+7-1=5$$

Simplify. The value x=\sqrt{13}+7 does not satisfy the equation.

$$2\times 13^{\frac{1}{2}}+7=5$$

Equation \sqrt{2x}=6-x has a unique solution.

$$x=7-\sqrt{13}$$