What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$

Question

Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?

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This is what I did: I try to multiply by the conjugate. Its value I believe is technically the solution. $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=24$. Given that $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3$, $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})= 3(\sqrt {49-x^2} + \sqrt {25-x^2}) =24\implies \sqrt {49-x^2} + \sqrt {25-x^2}=8$

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My question is that will this method work for similar problems, and is there a faster method? Thanks!

Answer 1

You basically used the formula $$a+b=\frac{a^2-b^2}{a-b}$$ in an indirect way.

The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.

And for the same reason, the same trick can be used to calculate $$\sqrt{\mbox{nice}}\pm \sqrt{\mbox{nice}}$$ whenever when $\sqrt{\mbox{nice}}\mp \sqrt{\mbox{nice}}$ is given.

Answer 2

Multiplying both sides by $\sqrt {49-x^2} + \sqrt {25-x^2}$, yields:

$$(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=3(\sqrt {49-x^2} + \sqrt {25-x^2})$$

$$(49-x^2 - (25-x^2)=3(\sqrt {49-x^2} + \sqrt {25-x^2})$$

Thus,

$$\sqrt {49-x^2} + \sqrt {25-x^2}=8.$$

Answer 3

Another way: $$1) \ \sqrt{49-x^2}=\sqrt{25-x^2}+3 \Rightarrow \\ 49-x^2=25-x^2+9+6\sqrt{25-x^2} \Rightarrow \\ 2.5=\sqrt{25-x^2};\\ 2) \ \sqrt{25-x^2}=\sqrt{49-x^2}-3 \Rightarrow \\ 25-x^2=49-x^2+9-6\sqrt{49-x^2} \Rightarrow \\ \sqrt{49-x^2}=5.5.$$ Hence: $$\sqrt{49-x^2}+\sqrt{25-x^2}=5.5+2.5=8.$$