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There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?

$2\sin^2{x} -5\cos{x} -4 =0 $

Here’s what I did:

$2\sin^2{x} -5\cos{x} -4 =0 $

$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$

$2 \cos^2{x} + 5 \cos{x} + 2 =0$

This is a quadratic function of $\cos x$, thus,

$\cos{x} = (-5 + 3)/4$ or $\cos{x}= (-5 - 3)/4$

But, the answer given is $\cos{x}=-\frac{1}{2} $ and WolframAlpha says the same but doesn’t show how to do it.

What did I do wrong?

Update: Thank you very much, everyone. Turns out that I wrote the squareroot of 9 as squareroot of 3. My bad

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  • $\begingroup$ Check the quadratc formula for $ax^2+bx+c=0$, that is $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ $\endgroup$
    – user
    Jul 17 '18 at 16:34
  • $\begingroup$ Your original question had $\sqrt{3}$s in the results from the Quadratic Formula; these should have been $3$s. Your edit has placed the $3$s, which resolves that error. So ... Is your question "Why is there only one solution, when the Quadratic Formula gives me two?" (If so, simply check that only one of the results is a valid value for cosine.) $\endgroup$
    – Blue
    Jul 17 '18 at 16:40
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    $\begingroup$ Yes I’ve made a mistake myself, even on my paper. I’ve cleared the question. Thank you for your help! $\endgroup$
    – Magician
    Jul 17 '18 at 16:41
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To solve $2y^2+5y+2=0$ either note $$y=\frac {-5\pm \sqrt {25-16}}4$$ so that $y=-2$ or $y=-\frac 12$, and only the second qualifies as a possible value for a cosine, or note the factorisation $$2y^2+5y+2=(2y+1)(y+2)$$

Somehow you have ended up with $\sqrt 3$ rather than $3$ in your computation of the quadratic formula.

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  • $\begingroup$ Thank you very much. Yes, sorry! That was an error made by me $\endgroup$
    – Magician
    Jul 17 '18 at 16:37
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Simple factorization..

$$2 \cos^2 x + 5 \cos x + 2 =0$$ $$(2 \cos^2 x + 4 \cos x) +(\cos x + 2) =0$$ $$2 \cos x(\cos x +2)+( \cos x + 2) =0$$ $$(\cos x +2)( 2\cos x + 1) =0$$ $$\implies ( 2\cos x + 1) =0$$ $$\implies \cos x =-\frac 12$$

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    $\begingroup$ Thank you very much, very convenient $\endgroup$
    – Magician
    Jul 17 '18 at 16:36
  • $\begingroup$ yw @Magician ......... $\endgroup$
    – MtGlasser
    Jul 17 '18 at 16:37
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Yes from here (you are right) we have:

$$2\sin^2x -5\cos x -4 =0 \implies 2\cos^2 x+5\cos x +2=0$$

that is by quadratic formula (here was your mistake):

$$\cos x = \frac{-5\pm \sqrt{25-16}}{4}= \frac{-5\pm 3}{4}\implies \cos x=-\frac12$$

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  • $\begingroup$ Oh my, oh my. Thank you so much. I mistakenly replaced 3 with the square root of 3. Thank you very much $\endgroup$
    – Magician
    Jul 17 '18 at 16:36

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