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I'm new to Quadratic Equations and I am following the two formulas to find $x_1$ and $x_2$ and they are:

$\Delta = b^2 - 4ac$

$x = \dfrac{-(b) \pm \sqrt\Delta}{2a}$

but delta always gives a negative number;

for example I have the following equation:

$6x^2 + 11x - 35 = 0$

What I did was:

1. Calculate the Delta ($\Delta$) by using the formula above $11^2 - 4ac$

$11^2 = 121$ and $4ac = -840 = 4\times6\times(-35)$

$\Delta = 121 - 840 = 719$

and got $-719$ as result

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  • $\begingroup$ You have $\Delta=719$, where did the $-719$ come from? $\endgroup$ – lioness99a Jun 19 '17 at 21:01
  • $\begingroup$ Careful! The formula says $b^2-4ac$ in the discriminant, and $4ac = -840$, so you should in fact be adding! $\endgroup$ – Sean Roberson Jun 19 '17 at 21:01
  • $\begingroup$ @SeanRoberson what do u mean I should be adding ? $\endgroup$ – Feelsbadman Jun 19 '17 at 21:03
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    $\begingroup$ Watch: $121-(-840) = 121 + 840$. $\endgroup$ – Sean Roberson Jun 19 '17 at 21:11
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You have e mistake in the sign: $$ -4 ac = -4\cdot(6)\cdot(-35)=+840 $$

so: $$ \Delta =b^2-4ac=b^2+(-4ac)= 121+840=961=31^2 $$

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The discriminant $\Delta$ comes in three flavors.

  1. $\Delta > 0$: two real roots
  2. $\Delta = 0$: degenerate (repeated root)
  3. $\Delta < 0$: roots are complex conjugates $(x\pm iy)$

enter image description here


Here is a plot of the function in question:

problem

This plot has two distinct, real roots. Therefore, $\Delta > 0$.

To double check your work, note that $$ 6 x^2+11 x-35 = (2 x+7) (3 x-5) $$ so the roots are $$ x = -\frac{7}{2}, \qquad x = \frac{5}{3} $$

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$\Delta$ is not always negative, but sometimes it will be -- then the equation has no real solutions.

E.g.,:

  • $x^2 - x + 1 = 0 \implies \Delta = 1^2-4 = -3 < 0$ has no real solutions
  • $x^2 - 2x + 1 = 0 \implies \Delta = 2^2-4 = 0$ has exactly 1 real solution at $x=1$
  • $2x^2 - 3x + 1 = 0 \implies \Delta = 3^2-4\times 2 = 1 > 0$ has 2 real solutions, $x=1$ and $x=1/2$.
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  • $\begingroup$ but I always get a negative number when its __NUM__$x^2$ $\endgroup$ – Feelsbadman Jun 19 '17 at 21:02
  • $\begingroup$ @IdontReallywolf see the last bullet $\endgroup$ – gt6989b Jun 19 '17 at 21:06
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The quadratic $6x^2 + 11x - 35$ has a=$6$, b=$11$ and c=$-35$. Therefore the discriminant $\Delta=b^2-4ac=(11)^2-4 \cdot 6 \cdot (-35)=121-(-840)=961$.

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$6x^2+11x−35=0$

$a =6; b= 11; c= -35$

$b^2 = 121$.

$4ac = 4*6*(-35) =-840$

$b^2 - 4ac = 121 - (-840) = 121 + 841 = 961=31$.

So $\Delta = 961=31^2 > 0$ and $\sqrt{961} = 31$

$x = \frac {-11 \pm 31}{2*6}$.

$x = \frac {-42}{12}$ or $x = \frac {20}{12}$

$x = -\frac 72$ or $x = \frac 53$

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Ok, let's start in dealing with your mistake and then talking more generally about what a negative descriminant ($\Delta$) means.
In your case: $b^2 = 121$, $4ac=-840$ so $b^2-4ac=(121)-(-840)=+961$ (since minus a minus is plus).
Now, there will be cases when your descriminant is negative, this means there are "no real solutions", you will learn about "non-real solutions" later but the case of a negative descriminant is easy to picture graphically. Think of solving a quadratic equation as looking for where a parabola crosses the x axis (finding the roots of the parabola), there will be parabolas that never cross the x axis, those are the cases when the descriminant is negative.

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