Mastering the Quadratic Formula and Discriminant for Exam Success
Unlock the power of the quadratic formula to solve any quadratic equation, and use the discriminant to determine the number of real solutions without solving. Essential for higher tier mathematics.
What the examiner is testing
The examiner assesses your ability to accurately substitute values into the quadratic formula and correctly interpret the sign of the discriminant. Marks are awarded for setting up the formula, accurate calculation of the solutions, and for using the discriminant to state the number of real roots.
The method
- Rearrange the quadratic equation into the standard form \(ax^2 + bx + c = 0\).
- Identify the values of \(a\), \(b\), and \(c\), paying close attention to their signs.
- Calculate the discriminant, \(\Delta = b^2 - 4ac\).
- If \(\Delta > 0\), there are two distinct real solutions. Substitute \(a\), \(b\), and \(c\) into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) and calculate both values.
- If \(\Delta = 0\), there is one repeated real solution. Substitute \(a\), \(b\), and \(c\) into \(x = \frac{-b}{2a}\) (or the full formula, noting \(\sqrt{0}=0\)).
- If \(\Delta < 0\), there are no real solutions. State this clearly.
Worked example
Solve \(3x^2 + 5x = 2\) to 2 decimal places.
$$
\begin{aligned}
3x^2 + 5x &= 2 \\
3x^2 + 5x - 2 &= 0 \\
a &= 3, \quad b = 5, \quad c = -2 \\
\Delta &= b^2 - 4ac \\
&= (5)^2 - 4(3)(-2) \\
&= 25 - (-24) \\
&= 25 + 24 \\
&= 49 \\
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
x &= \frac{-(5) \pm \sqrt{49}}{2(3)} \\
x &= \frac{-5 \pm 7}{6} \\
x_1 &= \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \approx 0.33 \\
x_2 &= \frac{-5 - 7}{6} = \frac{-12}{6} = -2
\end{aligned}
$$
The solutions are \(x = 0.33\) (2 d.p.) and \(x = -2\).
Sanity check: If \(x = 1/3\), \(3(1/3)^2 + 5(1/3) = 3(1/9) + 5/3 = 1/3 + 5/3 = 6/3 = 2\). Correct.
If \(x = -2\), \(3(-2)^2 + 5(-2) = 3(4) - 10 = 12 - 10 = 2\). Correct.
Worked example: a harder one
Find the range of values for \(k\) for which the equation \(2x^2 - (k+1)x + 8 = 0\) has no real solutions.
The obvious first move is to try and solve for \(x\) directly using the quadratic formula. However, this won't work because \(k\) is an unknown, and the question asks for a range of values for \(k\), not specific solutions for \(x\). The phrase "no real solutions" is the key.
$$
\begin{aligned}
2x^2 - (k+1)x + 8 &= 0 \\
a &= 2, \quad b = -(k+1), \quad c = 8
\end{aligned}
$$
For no real solutions, the discriminant must be less than zero: \(\Delta < 0\).
$$
\begin{aligned}
b^2 - 4ac &< 0 \\
(-(k+1))^2 - 4(2)(8) &< 0 \\
(k+1)^2 - 64 &< 0 \\
(k+1)^2 &< 64
\end{aligned}
$$
Taking the square root of both sides:
$$
\begin{aligned}
\sqrt{(k+1)^2} &< \sqrt{64} \\
|k+1| &< 8
\end{aligned}
$$
This inequality means that \((k+1)\) must be between \(-8\) and \(8\).
$$
\begin{aligned}
-8 &< k+1 < 8 \\
-8 - 1 &< k < 8 - 1 \\
-9 &< k < 7
\end{aligned}
$$
The range of values for \(k\) for which the equation has no real solutions is \(-9 < k < 7\).
Practice
- Solve \(x^2 + 7x + 10 = 0\).
- Solve \(5x^2 - 2x - 4 = 0\), giving your answers to 3 significant figures.
- Find the number of real solutions for the equation \(4x^2 - 12x + 9 = 0\).
- The equation \(x^2 + (p-3)x + p = 0\) has exactly one real solution. Find the possible values of \(p\).
Answers:
1. \(x = -2\), \(x = -5\)
2. \(x = 1.08\), \(x = -0.679\) (3 s.f.)
3. One real solution (since \(\Delta = 0\))
4. For exactly one real solution, \(\Delta = 0\).
$$
\begin{aligned}
a &= 1, \quad b = (p-3), \quad c = p \\
b^2 - 4ac &= 0 \\
(p-3)^2 - 4(1)(p) &= 0 \\
(p^2 - 6p + 9) - 4p &= 0 \\
p^2 - 10p + 9 &= 0 \\
(p-1)(p-9) &= 0 \\
p &= 1 \quad \text{or} \quad p = 9
\end{aligned}
$$
The possible values of \(p\) are \(1\) and \(9\).
The three mistakes that lose marks
- Incorrectly identifying \(a\), \(b\), or \(c\): Often happens when the equation isn't in standard form or when dealing with negative signs.
- Wrong answer produced: For \(x^2 - 3x = 5\), taking \(c=5\) instead of \(c=-5\). This leads to \(x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(5)}}{2(1)} = \frac{3 \pm \sqrt{9 - 20}}{2} = \frac{3 \pm \sqrt{-11}}{2}\), resulting in no real solutions when there should be two.
- Sign errors in the formula: Especially with \(-b\) or \(-4ac\).
- Wrong answer produced: For \(x^2 + 5x + 6 = 0\), if \(-b\) is written as \(5\) instead of \(-5\), then \(x = \frac{5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{1}}{2}\), giving \(x=3\) and \(x=2\), when the correct answers are \(x=-2\) and \(x=-3\).
- Calculation errors with \(\sqrt{b^2 - 4ac}\): Often \((-b)^2\) is incorrectly calculated as negative.
- Wrong answer produced: For \(x^2 - 5x + 6 = 0\), if \((-5)^2\) is calculated as \(-25\) instead of \(25\), then \(x = \frac{5 \pm \sqrt{-25 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{-49}}{2}\), leading to no real solutions when there should be two.
30-second recap
Rearrange to \(ax^2+bx+c=0\). Use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for solutions. The discriminant \(\Delta = b^2 - 4ac\) tells you the number of real solutions: \(\Delta > 0\) (two), \(\Delta = 0\) (one), \(\Delta < 0\) (none).