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We were putting equations into vertex form to factor as part of solving, and we were told to use the equation $c=\left(\frac{b}{2}\right)^2$. I figured you could skip vertex form and factor more easily using $c=b^2$. These two solving methods gave different results for x.

$$x^2+5x+3=0\\x^2+5x+c=-3+c\\x^2+5x+\frac{25}{4}=\frac{37}{4}\\\left(x+\frac{5}{2}\right)^2=\frac{37}{4}\\x+\frac{5}{2}=\sqrt{\frac{37}{4}}\\x=\frac{5}{2}\pm\frac{i\sqrt{37}}{2}\\x=5\pm i\sqrt{37}$$

As you can see, this is quite a complicated method of solving. Why don't we just use $b^2$:

$$x^2+5x+3=0\\x^2+5x+c=-3+c\\x^2+5x+25=22\\(x+5)^2=22\\x+5=\sqrt{22}\\x=5\pm\sqrt{22}$$

but now we have different results? Why is this? Is $c=b^2$ invalid? Did I make a mistake? If I'm adding $b^2$ to both sides, it should work, shouldn't it?

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  • $\begingroup$ @MyGlasses wow, stupid mistake. Can't believe I spent so much time writing this question $\endgroup$ – Travis Jan 18 '17 at 16:56
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    $\begingroup$ Dear downvoter: While the solution is trivial, I made an effort to show my work, and posted an actual problem I have had, is there a specific reason you downvoted? $\endgroup$ – Travis Jan 18 '17 at 17:02
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Both solutions have mistakes.

In the first one, the $\dfrac{37}4$ is not correct. It should be: $$ -3 + \frac{25}4 = -\frac{12}4 + \frac{25}4 = \frac{13}4$$

Also, the second and third lines below are incorrect aside from the $37/4$:

\begin{align*} \left(x + \frac52\right)^2 &= \frac{37}4\\[0.3cm] x + \frac52 &= \sqrt\frac{37}4\\[0.3cm] x &= \frac52 + i\frac{\sqrt{37}}2 \end{align*}

In the first line, just replace $37/4$ with $13/4$. Then the second line there should be: $$ x + \frac52 = \pm \sqrt{\frac{13}4} $$ and the third line should be: $$ x = - \frac52 \pm \frac{\sqrt{13}}2$$ And then that's it for that method.

For the second solution process, I don't know what's going on but it looks like you're just trying to complete the square again. There's only one way to do that, and that's what the first solution process is.

If you want to do a different method entirely, use the quadratic formula:

\begin{align*} x^2 + 5x + 3 &= 0\\ x &= \frac{-5 \pm \sqrt{5^2 - 4(1)(3)}}{2(1)}\\[0.3cm] x &= \frac{-5 \pm \sqrt{25 - 12}}2\\[0.3cm] x &= \frac{-5 \pm \sqrt{13}}2\\[0.3cm] x &= -\frac52 \pm \frac{\sqrt{13}}2 \end{align*}

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$$ (x + 5)^2 ≠ x^2 + 5x + 25. $$

Except for $x = 0$.

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$(x+5)^2=x^2+10x+25$ and not $x^2+5x+25$

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In first method you have mistake.

$-3 + \frac{25}4 = \frac{13}4 ≠ \frac{37}4$

In second method -

$x^2 + 5x + 25 ≠ (x + 5)^2$

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