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

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

Solve for m

$$m=10$$

Solution Steps

Subtract 5 from both sides of the equation.

$$\sqrt{m-1}=m-2-5$$

Subtract 5 from -2 to get -7.

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

Square both sides of the equation.

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

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

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

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

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

Subtract m^{2} from both sides.

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

Add 14m to both sides.

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

Combine m and 14m to get 15m.

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

Subtract 49 from both sides.

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

Subtract 49 from -1 to get -50.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -m^{2}+am+bm-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 -m^{2}+15m-50 as \left(-m^{2}+10m\right)+\left(5m-50\right).

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

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

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

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

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

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

$$m=10 m=5$$

Substitute 10 for m in the equation \sqrt{m-1}+5=m-2.

$$\sqrt{10-1}+5=10-2$$

Simplify. The value m=10 satisfies the equation.

$$8=8$$

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

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

Simplify. The value m=5 does not satisfy the equation.

$$7=3$$

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

$$m=10$$