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Solve for b in terms of a,x and c in the fastest way possible.

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closed as off-topic by JonMark Perry, Glorfindel, SteveV, rhsquared, Deusovi Jan 7 at 21:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – JonMark Perry, Glorfindel, SteveV, rhsquared, Deusovi
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ "subject" is a somewhat vague term. What you're presumably asking for is known in math as "isolating" or "solving for". "fastest" is also ambiguous. Do you mean "fewest operations"? $\endgroup$ – Acccumulation Jan 7 at 20:43
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From the quadratic equation

we can go back to the original equation $$ ax^2 + bx + c = 0 $$ and solve it for $b$, which yields: $$ b = - \frac{ax^2 + c}{x} $$

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  • $\begingroup$ Wow, I know mathematics university graduates who couldn't crack that one out. I love this exchange. Speaking of which. How do the tickings work for people giving out questions with both 1 answer, and an answer the OP knows? $\endgroup$ – yolo Jan 7 at 20:33
  • $\begingroup$ If you know an even faster way, then wait some time, so that people have a chance to come up with it on their own. $\endgroup$ – A. P. Jan 7 at 20:34
  • $\begingroup$ And since this is the fastest way I know of? $\endgroup$ – yolo Jan 7 at 20:35
  • $\begingroup$ @yolo Then you may tick it as soon as you want. But I fear that this question is not 'puzzling' enough and might get closed. But you could also wait whether someone comes up with a more puzzling-like solution. $\endgroup$ – A. P. Jan 7 at 20:37
  • $\begingroup$ A.P I think the firsy sentence in your answer is a bit of a giveaway... $\endgroup$ – yolo Jan 7 at 20:48
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Not the fastest... I kind of assumed we had to pretend we didn't already know the answer.

Multiply by 2a

$$2ax=-b \pm \sqrt{b^2-4ac}$$

Square everything

$$2ax*2ax=b^2+b^2-4ac \pm 2b\sqrt{b^2-4ac}$$

Simplify

$$4a^2x^2=-4ac-2b(-b\pm\sqrt{b^2-4ac})$$

Hey that looks familiar

$$4a^2x^2=-4ac-4bax$$

Finish it up

$$b=-\frac{ax^2+c}{x}$$

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  • $\begingroup$ 'we had to pretend we didn't already know the answer'- how so? I've never known anyone who knows the formula for b as general knowledge. $\endgroup$ – yolo Jan 7 at 20:46
  • $\begingroup$ ax^2+bx+c=0 is common knowledge and pretty much is solved. That answer is correct; i think this adds to the discussion but is not fast enough. $\endgroup$ – kaine Jan 7 at 20:56

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