Solve (sqrt{a}+sqrt{b})(sqrt{a}-sqrt{b})=a-b

$$( \sqrt { a } + \sqrt { b } ) ( \sqrt { a } - \sqrt { b } ) = a - b$$

Solve for a

$$a\geq 0$$

$$b\geq 0$$

Solution Steps

Consider \left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate \sqrt{b} to the power of 2 and get b.

$$a-b=a-b$$

Subtract a from both sides.

$$a-b-a=-b$$

Combine a and -a to get 0.

$$-b=-b$$

Cancel out -1 on both sides.

$$b=b$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\geq 0$$

$$a\geq 0$$

Solution Steps

Consider \left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate \sqrt{b} to the power of 2 and get b.

$$a-b=a-b$$

Add b to both sides.

$$a-b+b=a$$

Combine -b and b to get 0.

$$a=a$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$