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I know how to solve for $k$ in $kx^2-30x+25=0$ using $b^2-4ac$, but i want to know how the discriminant does this. How are we able to just plug the coefficients into the discriminant and get the answer for $k$? How does it all work? Any general overview would be great.

Rustyn
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Adam
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2 Answers2

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The discriminant can't tell you the value of $k$. What it can do is tell you how many real roots the function will have. If the discriminant is positive, then it will have two distinct real roots; if it's $0$, it will have two real roots which are both the same, and if it's negative, then the function has no real roots.

In your case, $k$ is just a constant. You would need more information in order to determine its value.

To elaborate, look at the quadratic equation

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

Notice that if the discriminant is positive, then we are taking a positive square root of the number. The two real solutions will be what we get when we add the value and when we subtract it (the $ \pm$).

If it's $0$, then we can' t have two different solutions, because $a - 0 = a + 0$.

If it's negative, then clearly there are no real solutions, because the square root of a negative number is not real.

user28375028
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  • (i) The quadratic equation kx^2-4x+k=0 has equal roots. Find the possible values of k.

    Answer: sub into b^2-4ac

        (-4)^2-4*k*k
    16-4k^2=0
    k^2=4
    k=2 or k=-2
    – Adam Oct 01 '14 at 19:35
  • That question was taken from an exam. Sorry if the layout is a bit confusing. – Adam Oct 01 '14 at 19:39
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Let the two roots be $r_1$ and $r_2$. Then the discriminant is by definition $$D=a^2(r_1-r_2)^2$$

so by taking the square root you are finding $r_1-r_2$ the difference of the roots. If the equation is $ax^2+bx+c$ then $r_1+r_2=-\frac{b}{a}$ So once you know $r_1+r_2$ and $r_1-r_2=\frac{\sqrt{D}}{a}$, then you can find the roots,

$$r_1=\frac{(r_1+r_2)+(r_1-r_2)}{2}=\frac{-b+\sqrt{D}}{2a}$$ $$r_2=\frac{(r_1+r_2)-(r_1-r_2)}{2}=\frac{-b-\sqrt{D}}{2a}$$

I guess all that remains to say is how to find $D=a^2(r_1-r_2)^2$. Here we note that $r_1r_2=\frac{c}{a}$ and

$$D=a^2(r_1-r_2)^2=a^2[(r_1+r_2)^2-4r_1r_2]=b^2-4ac$$

You can now easily put this together to find the quadratic formula. In general the way to solve polynomial equations is to find expressions of the roots with some symmetry and find them by solving auxillary equations (resolvents). In this case the resolvent equation is easy, its just $y^2=D$.

Rene Schipperus
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