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