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Question: Can I solve

$$4x^2 + 7x - 2 = 0$$

using the quadratic formula?

Attempt:

I used $$x = \frac{-b \pm \sqrt {b^2-4ac} } {2a}.$$

My answers were $-4$ and $-11$.

Am I right?

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    $\begingroup$ Your answers are not correct. In order to better help you can you please list the steps you took to arrive at those answers so that we can help spot where you may have made a mistake. $\endgroup$
    – Mufasa
    Commented Feb 12, 2015 at 22:07
  • $\begingroup$ You can check the correctness of your answers with WolframAlpha $\endgroup$
    – user147263
    Commented Feb 12, 2015 at 22:09
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    $\begingroup$ And this being your 66th question here, it's not too soon to learn proper math formatting. $\endgroup$
    – user147263
    Commented Feb 12, 2015 at 22:10
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    $\begingroup$ The formula you used, apart from missing parentheses, is correct, and can certainly be used. We get $\frac{-7\pm\sqrt{81}}{8}$, which can be simplified. $\endgroup$ Commented Feb 12, 2015 at 22:15
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    $\begingroup$ I, for one, would appreciate it if you could explain how you arrived at your answers. Could you show your work? $\endgroup$
    – mweiss
    Commented Feb 12, 2015 at 23:46

3 Answers 3

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Your formula is correct. Substituting in $a=4$, $b=7$ and $c=-2$ we have $$x=\dfrac{-7 \pm \sqrt{7^2 - 4(4)(-2)}}{2(4)} = \dfrac{-7 \pm \sqrt{49 - (-32)}}{8}$$

Can you finish from here?

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  • $\begingroup$ No need to do that to check that the formula is correct. It suffice to calculate directly $4(-4)^2 +7(-4)-2$ and $4(-11)^2 +7(-11)-2$ and to check that both are equal to zero. $\endgroup$
    – Olórin
    Commented Feb 12, 2015 at 22:28
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    $\begingroup$ @RobertGreen The comments on the question have already pointed out the poster's answers weren't correct - the answer was to confirm the quadratic formula they're using is correct and a prod in the right direction. :) $\endgroup$ Commented Feb 12, 2015 at 22:31
  • $\begingroup$ Ah ok, I thought the question was simply "Am I right?" and the answer "no", because $4(-4)^2 +7(-4)-2= 34$ for instance. $\endgroup$
    – Olórin
    Commented Feb 12, 2015 at 22:34
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Yes, you may use quadratic formula, and no, you didn't get it right.

By completing the square

$$\begin{align} 0 = 4 x^2 + 7x - 2 &= 4\Big(x^2 + \frac{7}{4}x\Big) - 2 \Rightarrow 4\Bigg[\Big(x + \frac{7}{8}\Big)^2 - \frac{49}{64}\Bigg] - 2 \\&\Rightarrow 4\Big(x + \frac{7}{8}\Big)^2 = \frac{49 + 32}{16} \\&\Rightarrow x = \frac{-7 \pm 9}{8}\\&\Rightarrow x =\begin{cases} \frac{1}{4}\\-2\end{cases}\end{align}$$

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Personally, I try not to invoke the quadratic formula unless I really need to. This usually means I use it when I think I will get complex or irrational roots. In your case, simply factoring will do the trick.

The coefficient of $x^2$ is $4$ and the constant term is $-2$. Also, the product of $4$ and $-2$ is $8$. Now, the factors of $-8$ which sum to $7$ are $-1$ and $8$. Thus, $$ 4x^2+7x-2 = 4x^2+8x-x-2=2(4x-1)+x(4x-1). $$ When we combine like terms, we get $$ (2+x)(4x-1). $$ Now, you wanted to figure out when $$ 4x^2+7x-2 = (2+x)(4x-1) = 0. $$ The answers are now pretty clear: $x=-2$ and $x=\frac{1}{4}$.


Of course, your question title asks whether or not you can use the quadratic formula, but, as seen above, it is really unnecessary (i.e., "overkill") to use it in this particular case.

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    $\begingroup$ The root $x = \frac{1}{4}$ is rational, but it is not an integer. Did you mean that you only use the quadratic formula when you think you will get complex or irrational roots? $\endgroup$ Commented Feb 12, 2015 at 23:14
  • $\begingroup$ @N.F.Taussig That is exactly what I meant--sorry for the brain fart. :) $\endgroup$ Commented Feb 12, 2015 at 23:15

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