Solve sqrt{x-1}=x-7

$$\sqrt{ x-1 } = x-7$$

Solve for x

$$x=10$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x-1}\right)^{2}=\left(x-7\right)^{2}$$

Calculate \sqrt{x-1} to the power of 2 and get x-1.

$$x-1=\left(x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x-1=x^{2}-14x+49$$

Subtract x^{2} from both sides.

$$x-1-x^{2}=-14x+49$$

Add 14x to both sides.

$$x-1-x^{2}+14x=49$$

Combine x and 14x to get 15x.

$$15x-1-x^{2}=49$$

Subtract 49 from both sides.

$$15x-1-x^{2}-49=0$$

Subtract 49 from -1 to get -50.

$$15x-50-x^{2}=0$$

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

$$-x^{2}+15x-50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-50. To find a and b, set up a system to be solved.

$$a+b=15 ab=-\left(-50\right)=50$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 50.

$$1,50 2,25 5,10$$

Calculate the sum for each pair.

$$1+50=51 2+25=27 5+10=15$$

The solution is the pair that gives sum 15.

$$a=10 b=5$$

Rewrite -x^{2}+15x-50 as \left(-x^{2}+10x\right)+\left(5x-50\right).

$$\left(-x^{2}+10x\right)+\left(5x-50\right)$$

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

$$-x\left(x-10\right)+5\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x+5\right)$$

To find equation solutions, solve x-10=0 and -x+5=0.

$$x=10 x=5$$

Substitute 10 for x in the equation \sqrt{x-1}=x-7.

$$\sqrt{10-1}=10-7$$

Simplify. The value x=10 satisfies the equation.

$$3=3$$

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

$$\sqrt{5-1}=5-7$$

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

$$2=-2$$

Equation \sqrt{x-1}=x-7 has a unique solution.

$$x=10$$