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The question goes on about like this.

$$-2 = \sqrt{7 -2 b} - \sqrt{2 b + 3}$$

I tried squaring both sides but than that does not make sense as one side has 2 square roots. Then do I double it, like $(-2)^4$ 'cause there are two of them?

MATH ASKER
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2 Answers2

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Notice, we have $$-2=\sqrt{7-2b}-\sqrt{2b+3}$$ Squaring both the sides, we get $$(-2)^2=(\sqrt{7-2b}-\sqrt{2b+3})^2$$ $$4=7-2b+2b+3-2\sqrt{7-2b}\sqrt{2b+3}$$ $$\sqrt{7-2b}\sqrt{2b+3}=3$$ Again, squaring both the sides, we get $$(7-2b)(2b+3)=3^2$$ $$b^2-2b-3=0$$$$\implies (b+1)(b-3)=0$$

I hope you can solve further

Harish Chandra Rajpoot
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  • Thanks a lot for answering but how did you get cause if you square the other side dont you cancel the square root, why is it there – MATH ASKER Sep 30 '15 at 00:29
  • As you see in the answer, we square twice the equation to eliminate square root. You can ask further if you have any query – Harish Chandra Rajpoot Sep 30 '15 at 00:31
  • Thanks again for answering but how did you get here sqrt(7-2b)(2b+3) = 3 what happened to 4=7-2b+2b-2sqrt(7-2b)(sqrt(2b+3) – MATH ASKER Sep 30 '15 at 00:36
  • you see that $2b$ term is cancelled out then we have $LHS=7-2b+2b+3-2\sqrt{7-2b}\sqrt{2b+3}=10-2\sqrt{7-2b}\sqrt{2b+3}=2(5-\sqrt{7-2b}\sqrt{2b+3}$ Now, divide by $2$ take $\sqrt{7-2b}\sqrt{2b+3}$ to the right side & constant term to the left finally, you find $$\sqrt{7-2b}\sqrt{2b+3}=3$$ – Harish Chandra Rajpoot Sep 30 '15 at 00:39
  • Ohh did you move 2 and 5 to the other side? but how would you do that..to take it to the other side you would have to divide 2? i think i might be making this complex but would you be able to tell me how you got = 3 ?? – MATH ASKER Sep 30 '15 at 00:50
  • Yeah, of course. You are absolutely correct. See when you take $2$ & $5$ to the other (right) side you get $5-2=3$ – Harish Chandra Rajpoot Sep 30 '15 at 00:51
  • The solution $b=-1$ is a parasitc solution, due to squaring. – Bernard Sep 30 '15 at 01:06
  • yes, you are correct, $b=3$ is acceptable value – Harish Chandra Rajpoot Sep 30 '15 at 01:08
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You eliminate the radicals by squaring, repeating the operation as many times as necessary. The simplest, in general is to isolate a radical in one side.

However, beware the equation obtained after squaring is implied by the original equation, but not equivalent to it. Indeed the main rule is: $$A=\sqrt B\iff A^2=B\enspace \color{red}{\text{and}}\enspace A\ge 0$$ Here, we first observe the domain of validity of the equation is $\;\biggl[-\dfrac32,\dfrac72\biggr]$.

We can rewrite the equation as \begin{align*} \sqrt{7-2b}+2&=\sqrt{2b+3}\Rightarrow 7-2b+4+4\sqrt{7-2b}=2b+3\iff\sqrt{7-2b}=b-2\\&\iff 7-2b=b^2-4b+4\enspace \color{red}{\text{and}}\enspace b\ge 2\\ &\iff b^2-2b-3=(b+1)(b-3)=0\enspace \color{red}{\text{and}}\enspace b\ge 2 \end{align*} Thus there is only one solution: $\;\color{red}{b=3}$.

You can check the solution to the final equation, $b=-1$, is not a solution to the initial equation, as the right-hand side is equal to $+2$ for this value of $b$.

Bernard
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