Solve sqrt{a-4}+1=sqrt{2a-7}

$$\sqrt { a - 4 } + 1 = \sqrt { 2 a - 7 }$$

Solve for a

$$a=8$$

$$a=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{a-4}+1\right)^{2}=\left(\sqrt{2a-7}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{a-4}+1\right)^{2}.

$$\left(\sqrt{a-4}\right)^{2}+2\sqrt{a-4}+1=\left(\sqrt{2a-7}\right)^{2}$$

Calculate \sqrt{a-4} to the power of 2 and get a-4.

$$a-4+2\sqrt{a-4}+1=\left(\sqrt{2a-7}\right)^{2}$$

Add -4 and 1 to get -3.

$$a-3+2\sqrt{a-4}=\left(\sqrt{2a-7}\right)^{2}$$

Calculate \sqrt{2a-7} to the power of 2 and get 2a-7.

$$a-3+2\sqrt{a-4}=2a-7$$

Subtract a-3 from both sides of the equation.

$$2\sqrt{a-4}=2a-7-\left(a-3\right)$$

To find the opposite of a-3, find the opposite of each term.

$$2\sqrt{a-4}=2a-7-a+3$$

Combine 2a and -a to get a.

$$2\sqrt{a-4}=a-7+3$$

Add -7 and 3 to get -4.

$$2\sqrt{a-4}=a-4$$

Square both sides of the equation.

$$\left(2\sqrt{a-4}\right)^{2}=\left(a-4\right)^{2}$$

Expand \left(2\sqrt{a-4}\right)^{2}.

$$2^{2}\left(\sqrt{a-4}\right)^{2}=\left(a-4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{a-4}\right)^{2}=\left(a-4\right)^{2}$$

Calculate \sqrt{a-4} to the power of 2 and get a-4.

$$4\left(a-4\right)=\left(a-4\right)^{2}$$

Use the distributive property to multiply 4 by a-4.

$$4a-16=\left(a-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-4\right)^{2}.

$$4a-16=a^{2}-8a+16$$

Subtract a^{2} from both sides.

$$4a-16-a^{2}=-8a+16$$

Add 8a to both sides.

$$4a-16-a^{2}+8a=16$$

Combine 4a and 8a to get 12a.

$$12a-16-a^{2}=16$$

Subtract 16 from both sides.

$$12a-16-a^{2}-16=0$$

Subtract 16 from -16 to get -32.

$$12a-32-a^{2}=0$$

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

$$-a^{2}+12a-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-32. To find a and b, set up a system to be solved.

$$a+b=12 ab=-\left(-32\right)=32$$

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 32.

$$1,32 2,16 4,8$$

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 12.

$$a=8 b=4$$

Rewrite -a^{2}+12a-32 as \left(-a^{2}+8a\right)+\left(4a-32\right).

$$\left(-a^{2}+8a\right)+\left(4a-32\right)$$

Factor out -a in the first and 4 in the second group.

$$-a\left(a-8\right)+4\left(a-8\right)$$

Factor out common term a-8 by using distributive property.

$$\left(a-8\right)\left(-a+4\right)$$

To find equation solutions, solve a-8=0 and -a+4=0.

$$a=8 a=4$$

Substitute 8 for a in the equation \sqrt{a-4}+1=\sqrt{2a-7}.

$$\sqrt{8-4}+1=\sqrt{2\times 8-7}$$

Simplify. The value a=8 satisfies the equation.

$$3=3$$

Substitute 4 for a in the equation \sqrt{a-4}+1=\sqrt{2a-7}.

$$\sqrt{4-4}+1=\sqrt{2\times 4-7}$$

Simplify. The value a=4 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{a-4}+1=\sqrt{2a-7}.

$$a=8 a=4$$