This may sound a bit basic but when using the quadratic formula working through $\sqrt{b^2- 4ac}$, should I be considering $- 4ac $ or $-(4ac) $? Not sure if this is worded correctly but some feedback would be helpful!

  • $\begingroup$ no it is $$\sqrt{b^2-4ac}$$ $\endgroup$ – Dr. Sonnhard Graubner May 20 '17 at 13:38
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    $\begingroup$ Either way, $-4ac=-(4ac)$. $\endgroup$ – Wuestenfux May 20 '17 at 13:43

Either way it would be correct, in fact they are the same. If you a have an equation of the form $$ax^2 + bx + c = 0$$ and want to find $x$ then, as the formula shows, you get $$x = \frac{-b \pm \sqrt{b^2 - 4 ac}}{2a}$$Note that all these constants $a,b,c$ must have their actual sign. For example, consider $$ - 4x^2 + 2x + 3 = 0$$ then $$x = \frac{-2 \pm \sqrt{4 - 4\cdot(-4) \cdot3}}{2\cdot (-4)} = \frac{-2 \pm \sqrt{52}}{-8}$$

  • $\begingroup$ Im not sure what i did wrong on this problem that makes sense. Think its time to sleep! Thank you $\endgroup$ – Ryan May 20 '17 at 14:01

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