Solve sqrt{a+4}+sqrt{a-3}=7

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

Solve for a

$$a=12$$

Solution Steps

Subtract \sqrt{a-3} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-\sqrt{a-3}\right)^{2}.

$$a+4=49-14\sqrt{a-3}+\left(\sqrt{a-3}\right)^{2}$$

Calculate \sqrt{a-3} to the power of 2 and get a-3.

$$a+4=49-14\sqrt{a-3}+a-3$$

Subtract 3 from 49 to get 46.

$$a+4=46-14\sqrt{a-3}+a$$

Add 14\sqrt{a-3} to both sides.

$$a+4+14\sqrt{a-3}=46+a$$

Subtract a from both sides.

$$a+4+14\sqrt{a-3}-a=46$$

Combine a and -a to get 0.

$$4+14\sqrt{a-3}=46$$

Subtract 4 from both sides.

$$14\sqrt{a-3}=46-4$$

Subtract 4 from 46 to get 42.

$$14\sqrt{a-3}=42$$

Divide both sides by 14.

$$\sqrt{a-3}=\frac{42}{14}$$

Divide 42 by 14 to get 3.

$$\sqrt{a-3}=3$$

Square both sides of the equation.

$$a-3=9$$

Add 3 to both sides of the equation.

$$a-3-\left(-3\right)=9-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$a=9-\left(-3\right)$$

Subtract -3 from 9.

$$a=12$$

Substitute 12 for a in the equation \sqrt{a+4}+\sqrt{a-3}=7.

$$\sqrt{12+4}+\sqrt{12-3}=7$$

Simplify. The value a=12 satisfies the equation.

$$7=7$$

Equation \sqrt{a+4}=-\sqrt{a-3}+7 has a unique solution.

$$a=12$$