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In Higher Algebra by Hall and Knight, the following "artifice" for solving a certain type of equations is given:

Solve: $\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x-11} = 9$

They make use of the fact that $(3x^2-4x+34) - (3x^2-4x-11) = 45$, and utilizing the formula $a^2 - b^2 = (a-b)(a+b)$, obtain a neat answer that $x = 3, -5/3$.

My attempt, however, results in something weird:

$\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x+11} = 9$

Let $3x^2 - 4x = t$

$\Rightarrow \sqrt{t+34} + \sqrt{t+11}$ = 9

Squaring both sides,

$t + 34 + t + 11 + 2 \sqrt{(t+34)(t+11)} = 81$

$\Rightarrow 18-t = \sqrt{(t+34)(t+11)}$

Squaring again,

$t^2 - 36t + 324 = t^2 + 45t + 374$

$\Rightarrow t = -50/81$

$\Rightarrow 3x^2 - 4x = -50/81$

$\Rightarrow 243x^2 - 324x + 50 = 0$

The discriminant here is $\sqrt{324^2 - 200.243} = 56376$, which is not a perfect square and hence the roots are irrational.

Have I made a mistake somewhere or it is really the case that finesse leads to correct answers?

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In general, try to avoid squaring wherever practicable as it introduces extraneous root(s) which need(s) to be discarded after validation –  lab bhattacharjee Aug 17 '13 at 4:42
    
    
Unless that book's working under rather unusual agreements (e.g., taking sometimes the negative square root of positive real numbers), none of what you call "neat answers" is really a solution, as substituting any of them in the equations gives $$\sqrt{49}-\sqrt4=7-2=5\neq 9\;$$ Yet from what you wrote after your substitution $\,t=3x^2-4x\;$ it seems to be there a plus sign between both square roots...but then again in the first line the second sqrt has $\,-11\,$ but then it appears $\;+11\;$...so you have to be careful and write clearly what exactly is your eqaution! –  DonAntonio Aug 17 '13 at 8:26

2 Answers 2

up vote 1 down vote accepted

Oops! Realized that I had written $-11$ as $11$. Sorry for the mistake! I'll try again and re-post if needed.

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dotslash, also note that you wrote $\sqrt{t+34} + \sqrt{t+11}$, whereas it looks like you meant $\sqrt{t+34} - \sqrt{t+11}$. –  AWertheim Aug 17 '13 at 4:27
    
Exactly. Too many mistakes . . . I'm embarrassed! –  dotslash Aug 17 '13 at 4:31
    
dotslash, do not be embarrassed. We all make plenty of mistakes... they're good for us! And furthermore, I happen to like your approach a lot. Hall and Knight's trick here is very slick, but I think your approach is more illuminating. Keep up the good work :) –  AWertheim Aug 17 '13 at 4:33
    
@AWertheim Thank you! I just thought I wasted everyone's time carelessly. :( –  dotslash Aug 17 '13 at 4:53

Assuming the real, actual question has a plus sign between the square roots and $\,-11\,$ in the second square root, and doing your substitution:

$$\sqrt{3x^2-4x+34}+\sqrt{3x^2-4x-11}=9\;,\;\;\color{red}{t:=3x^2-4x}\implies$$

$$\sqrt{t+34}+\sqrt{t-11}=9\implies 2t+23+2\sqrt{(t+34)(t-11)}=81\implies$$

$$29-t=\sqrt{(t+34)(t-11)}\implies t^2-58t+841=t^2+23t-374\implies$$

$$81t=1,215\implies t=15\implies$$

$$15=\color{red}{t=3x^2-4x}\implies 0=3x^2-4x-15=3(x-3)\left(x+\frac53\right)$$

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