# Discriminant

The formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ gives the solutions to the general quadratic equation $ax^2+bx+c=0$. By examining the expression under the square root sign, $b^2-4ac$, we know as the discriminant symbol used is $\Delta$.

The quadratic formula becomes $x=\dfrac{-b + \sqrt{\Delta}}{2a}$ and $x=\dfrac{-b-\sqrt{\Delta}}{2a}$.

$\Delta <0$
If $x^2+2x+3=0$, then $a=1,b=2$ and $c=3$.
\begin{align} \displaystyle x &=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &= \dfrac{-2 \pm \sqrt{-8}}{2} \\ \Delta &= b^2-4ac \\ &= 2^2-4 \times 1 \times 3 \\ &= -8 \end{align}
There are no real solutions if the discriminant is less than zero (negative) because the expression under the square root sign is negative. Finding a real number, the square root of a negative number, is impossible.

$\Delta = 0$
If $x^2+8x+16=0$, then $a=1,b=8$ and $c=16$.
\begin{align} \displaystyle x &=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &= \dfrac{-8 \pm \sqrt{0}}{2}\\ &= -4 \\ \Delta &= b^2-4ac \\ &= 8^2-4 \times 1 \times 16 \\ &= 0 \end{align}
If the discriminant is equal to zero, then the two solutions are the same. This may be regarded as one rational solution equal to $\displaystyle -\dfrac{b}{2a}$.
This is, if $b^2-4ac=0$, then $\displaystyle x=\dfrac{-b+\sqrt{0}}{2a}$ which is the same as $\displaystyle x=\dfrac{-b-\sqrt{0}}{2a}$.
One solution indicates that the quadratic trinomial is a perfect square that can be factorised easily using the rule of the perfect square: that is $x^2+8x+16=(x+4)^2$.
We also call this repeated or double root.

$\Delta > 0$
If the discriminant is positive, there are two distinct solutions. We can determine more information than this by checking whether the discriminant is a perfect square.
Case 1:
If $2x^2-7x-4=0$, then $a=2,b=-7$ and $c=-4$.
\begin{align} \displaystyle x &=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &= \dfrac{7 \pm \sqrt{81}}{2 \times 2} \\ &= \dfrac{7 \pm 9}{4} \\ &= 4 \text{ or } x=-\dfrac{1}{2} \\ \Delta &= b^2-4ac \\ &= (-7)^2 -4 \times 2 \times (-4) \\ &= 81 \end{align}
If the discriminant is positive and has a perfect square, the quadratic trinomial will have two rational solutions. This means the quadratic trinomial can be factorised easily; that is $2x^2-7x-4=(2x+1)(x-4)$.
Case 2:
If $x^2-5x-1=0$ then $a=1,b=-5$ and $c=-1$.
\begin{align} \displaystyle x &=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &= \dfrac{5 \pm \sqrt{29}}{2 \times 1}\\ &= \dfrac{5 \pm \sqrt{29}}{2}\\ \Delta &= b^2-4ac \\ &= (-5)^2-4 \times 1 \times (-1) \\ &= 29 \end{align}
If the discriminant is positive but not a perfect square, the factors are irrational, and the quadratic formula is used to find the two irrational (surd) solutions.

### Example 1

Use the discriminant to determine the nature of the roots of $2x^2-2x+5=0$.

\begin{align} \displaystyle \Delta &= b^2-4ac \\ &= (-2)^2-4 \times 2 \times 5 \\ &= -36 \end{align}
Since $\Delta \lt 0$, there are no real roots.

### Example 2

Use the discriminant to determine the nature of the roots of $3x^2-4x-2=0$.

\begin{align} \displaystyle \Delta &= b^2-4ac \\ &= (-4)^2-4 \times 3 \times (-2) \\ &= 40 \end{align}
Since $\Delta \gt 0$, but $40$ is not a square, there are $2$ distinct irrational roots.

### Example 3

Use the discriminant to determine the nature of the roots of $9x^2-6x+1=0$.

\begin{align} \displaystyle \Delta &= b^2-4ac \\ &= (-6)^2 -4 \times 9 \times 1 \\ &= 0 \end{align}
Since $\Delta = 0$, the equation has a repeated root.

### Example 4

Find the values of $k$ for which $x^2-4x+k=0$ has two distinct real roots.

\begin{align} \displaystyle \Delta &\gt 0 \\ b^2-4ac &\gt 0 \\ (-4)^2-4 \times 1 \times k &\gt 0 \\ 16-4k &\gt 0\\ -4k &\gt-16 \\ \therefore k &\lt 4 \end{align}

### Example 5

Find the values of $k$ for which $kx^2+(k+3)x-1=0$ has real roots.

\begin{align} \displaystyle \Delta &\ge 0 &\text{for real roots} \\ b^2-4ac &\ge 0 \\ (k+3)^2-4 \times k \times (-1) &\ge 0 \\ k^2 + 6k + 9 + 4k &\ge 0\\ k^2 + 10k + 9 &\ge 0\\ (k+1)(k+9) &\ge 0 \\ \therefore k &\le -9 \text{ or } \ge -1 \end{align}