Solve sqrt{2x+35}=x

$$\sqrt { 2 x + 35 } = x$$

Solve for x

$$x=7$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+35}\right)^{2}=x^{2}$$

Calculate \sqrt{2x+35} to the power of 2 and get 2x+35.

$$2x+35=x^{2}$$

Subtract x^{2} from both sides.

$$2x+35-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+2x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=2 ab=-35=-35$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=7 b=-5$$

Rewrite -x^{2}+2x+35 as \left(-x^{2}+7x\right)+\left(-5x+35\right).

$$\left(-x^{2}+7x\right)+\left(-5x+35\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(x-7\right)-5\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x-5\right)$$

To find equation solutions, solve x-7=0 and -x-5=0.

$$x=7 x=-5$$

Substitute 7 for x in the equation \sqrt{2x+35}=x.

$$\sqrt{2\times 7+35}=7$$

Simplify. The value x=7 satisfies the equation.

$$7=7$$

Substitute -5 for x in the equation \sqrt{2x+35}=x.

$$\sqrt{2\left(-5\right)+35}=-5$$

Simplify. The value x=-5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-5$$

Equation \sqrt{2x+35}=x has a unique solution.

$$x=7$$