The discriminant

AS · P1 · 12 min

The discriminant $b^2 - 4ac$ tells you how many real roots a quadratic equation has without solving it. That makes it the key tool for questions with an unknown constant: "find $k$ so that the line is a tangent", "find the set of values of $m$ for which the expression is always positive". These appear on almost every Paper 1, usually worth 3 to 5 marks, and the setting-up is where most marks are won or lost.

Where it comes from

The quadratic formula is

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

Everything depends on the number under the square root.

If it is positive, the square root is a positive real number, and $+\sqrt{\phantom{x}}$ and $-\sqrt{\phantom{x}}$ give two different roots.
If it is zero, $\pm 0$ makes no difference, and both roots equal $-\dfrac{b}{2a}$ : a repeated root.
If it is negative, there is no real square root, so there are no real roots.

Definition

The discriminant of the quadratic $ax^2 + bx + c$ is $b^2 - 4ac$ . A repeated root (also called equal roots) is a root that occurs twice, as when the quadratic is a perfect square $a(x - \alpha)^2$ .

Key result

$b^2 - 4ac$	Roots of $ax^2 + bx + c = 0$	Graph of $y = ax^2 + bx + c$
$> 0$	two distinct real roots	crosses the $x$ -axis twice
$= 0$	one repeated real root, $x = -\dfrac{b}{2a}$	touches the $x$ -axis at its vertex
$< 0$	no real roots	does not meet the $x$ -axis

"Real roots" (with no other word) means $b^2 - 4ac \ge 0$ : either two distinct or one repeated.

y = (x - 2)^2 - 2 y = (x - 2)^2 y = (x - 2)^2 + 2

The three curves above differ only in their constant term. The lowest crosses the axis twice (discriminant positive), the middle one touches it (discriminant zero), and the highest misses it (discriminant negative).

Translating the wording

Exam questions rarely say "discriminant". They describe the situation, and you translate it.

The question says	Write
two distinct real roots; meets in two distinct points; cuts	$b^2 - 4ac > 0$
equal roots; a repeated root; is a tangent; touches	$b^2 - 4ac = 0$
no real roots; does not meet; does not intersect	$b^2 - 4ac < 0$
real roots; meets (not "in two distinct points")	$b^2 - 4ac \ge 0$
$ax^2 + bx + c$ is positive for all $x$	$a > 0$ and $b^2 - 4ac < 0$
$ax^2 + bx + c$ is negative for all $x$	$a < 0$ and $b^2 - 4ac < 0$

The last two rows need the sign of $a$ as well. A quadratic with no real roots lies entirely on one side of the $x$ -axis, and $a$ decides which side.

Discriminant questions with an unknown constant

Rearrange the equation to the form $ax^2 + bx + c = 0$ . If the problem involves a line and a curve, substitute first to get a single quadratic.
Identify $a$ , $b$ and $c$ , which may contain the unknown constant. Write them down.
Write the correct condition: $> 0$ , $= 0$ , $< 0$ or $\ge 0$ .
Simplify into an equation or inequality in the constant.
Solve it. An inequality in $k$ is itself usually a quadratic inequality: find the critical values and sketch.
Check whether the coefficient of $x^2$ could be zero for some value of the constant. That case needs separate thought.

Counting roots

Determine the number of real roots of (a) $2x^2 - 3x + 5 = 0$ and (b) $4x^2 - 12x + 9 = 0$ .

Solution

(a) $a = 2$ , $b = -3$ , $c = 5$ : $b^2 - 4ac = 9 - 40 = -31 < 0$ . No real roots.

(b) $a = 4$ , $b = -12$ , $c = 9$ : $b^2 - 4ac = 144 - 144 = 0$ . One repeated root (in fact $4x^2 - 12x + 9 = (2x - 3)^2$ , so the root is $x = \tfrac{3}{2}$ ).

Finding a constant for a repeated root

The equation $x^2 + kx + 16 = 0$ has a repeated root. Find the possible values of $k$ and, for each, the repeated root.

Solution

Repeated root means $b^2 - 4ac = 0$ :

k^2 - 4(1)(16) = 0 \quad\Rightarrow\quad k^2 = 64 \quad\Rightarrow\quad k = 8 \text{ or } k = -8

The repeated root is $x = -\dfrac{b}{2a} = -\dfrac{k}{2}$ .

$k = 8$ : $x^2 + 8x + 16 = (x + 4)^2$ , root $x = -4$ .
$k = -8$ : $x^2 - 8x + 16 = (x - 4)^2$ , root $x = 4$ .

Both values of $k$ are needed; giving only $k = 8$ loses a mark.

A set of values of k

Find the set of values of $k$ for which the equation $x^2 + (k - 3)x + 4 = 0$ has two distinct real roots.

Solution

$a = 1$ , $b = k - 3$ , $c = 4$ . Two distinct real roots:

(k - 3)^2 - 16 > 0

Critical values: $(k - 3)^2 = 16$ , so $k - 3 = \pm 4$ , giving $k = 7$ or $k = -1$ .

The expression $(k - 3)^2 - 16$ is a $\cup$ -shaped quadratic in $k$ , positive outside its roots:

y = (x - 3)^2 - 16 (-1, 0) (7, 0)

(The horizontal axis here is $k$ .) So $k < -1$ or $k > 7$ .

When the coefficient of x squared contains k

Find the set of values of $k$ for which the equation $kx^2 + 2kx + 3 = 0$ has no real roots.

Solution

Case $k \neq 0$ . The equation is quadratic with $a = k$ , $b = 2k$ , $c = 3$ . No real roots:

(2k)^2 - 4(k)(3) < 0 \quad\Rightarrow\quad 4k^2 - 12k < 0 \quad\Rightarrow\quad 4k(k - 3) < 0

The quadratic $4k(k - 3)$ is negative between its roots, so $0 < k < 3$ .

Case $k = 0$ . The equation becomes $0x^2 + 0x + 3 = 0$ , i.e. $3 = 0$ , which has no solutions at all. So $k = 0$ also gives no real roots.

Combining: $0 \le k < 3$ .

Most candidates write $0 < k < 3$ and lose the final mark. Whenever $a$ contains the constant, test the value that makes $a = 0$ .

Showing roots are always real

Show that the equation $px^2 + (p + 2)x + 1 = 0$ , where $p$ is a non-zero constant, has two distinct real roots for every value of $p$ .

Solution

$a = p$ , $b = p + 2$ , $c = 1$ :

b^2 - 4ac = (p + 2)^2 - 4p = p^2 + 4p + 4 - 4p = p^2 + 4

Since $p^2 \ge 0$ , we have $p^2 + 4 \ge 4 > 0$ for every $p$ . The discriminant is always positive, so there are always two distinct real roots.

For a "show that ... for all values" question, the discriminant usually simplifies to a completed square or a sum of squares plus a positive number. State clearly why it is positive.

A tangent to a curve

The line $y = 2x + k$ is a tangent to the curve $y = x^2 - 4x + 7$ . Find the value of $k$ and the coordinates of the point of contact.

Solution

At points of intersection, $x^2 - 4x + 7 = 2x + k$ , so

x^2 - 6x + (7 - k) = 0

A tangent meets the curve at exactly one point, so the discriminant is zero:

(-6)^2 - 4(1)(7 - k) = 0 \quad\Rightarrow\quad 36 - 28 + 4k = 0 \quad\Rightarrow\quad k = -2

Then $x^2 - 6x + 9 = 0$ , so $(x - 3)^2 = 0$ and $x = 3$ . Then $y = 2(3) - 2 = 4$ . The point of contact is $(3, 4)$ .

y = x^2 - 4x + 7 y = 2x - 2 (3, 4)

Negative for all x

Find the set of values of $m$ for which $-2x^2 + mx - 8 < 0$ for all real $x$ .

Solution

Here $a = -2 < 0$ , so the graph is $\cap$ -shaped. It is below the axis everywhere exactly when it has no real roots:

m^2 - 4(-2)(-8) < 0 \quad\Rightarrow\quad m^2 - 64 < 0 \quad\Rightarrow\quad (m - 8)(m + 8) < 0

So $-8 < m < 8$ .

Watch out

$k^2 < 64$ does not mean $k < 8$ . It means $-8 < k < 8$ . And $k^2 > 64$ means $k < -8$ or $k > 8$ . Always solve via the critical values and a sketch, never by "square rooting both sides" of an inequality.

Using the wrong condition. "Meets the curve" allows tangency, so it is $\ge 0$ . "Meets at two distinct points" is $> 0$ .

Not rearranging first. In $x^2 + 3x = kx - 1$ , the coefficient $b$ is $3 - k$ and $c$ is $1$ , not $b = 3$ , $c = 0$ .

Forgetting the case $a = 0$ . If the coefficient of $x^2$ contains $k$ , check separately what happens when it is zero.

Sign errors in $-4ac$ . With $a = -2$ and $c = -8$ , $-4ac = -4(-2)(-8) = -64$ . Put every coefficient in brackets.

Exam tip

Write $b^2 - 4ac$ with the numbers substituted in brackets before simplifying. That line alone usually earns a method mark.
State the condition in words or symbols ("for a tangent, $b^2 - 4ac = 0$ "). Examiners need to see that you know which condition applies.
For "set of values" answers, give the final answer as inequalities, for example $k < -1$ or $k > 7$ , or $-8 < m < 8$ . Writing " $-1 > k > 7$ " is wrong (it describes no numbers).
If a question asks for the point of contact of a tangent, find $k$ first, then solve the resulting perfect square.

Summary

Discriminant $= b^2 - 4ac$ : positive, two distinct roots; zero, a repeated root; negative, no real roots.
"Real roots" means $\ge 0$ . "Tangent" or "touches" means $= 0$ .
Always positive needs $a > 0$ and $b^2 - 4ac < 0$ ; always negative needs $a < 0$ and $b^2 - 4ac < 0$ .
Rearrange to $ax^2 + bx + c = 0$ and identify $a$ , $b$ , $c$ before substituting.
An inequality in $k$ is solved with critical values and a sketch.
If $a$ depends on $k$ , test the value of $k$ that makes $a = 0$ .
"Show that roots are real for all $k$ ": simplify the discriminant to something visibly positive, such as $k^2 + 4$ .

Practice questions

Question

Find the discriminant of $3x^2 - 5x + 4$ and hence state the number of real roots of $3x^2 - 5x + 4 = 0$ .
The equation $x^2 - 6x + c = 0$ has a repeated root. Find $c$ and the root.
Find the values of $k$ for which $4x^2 + kx + 9 = 0$ has equal roots.
Find the set of values of $k$ for which $x^2 + kx + k + 3 = 0$ has two distinct real roots.
Find the set of values of $p$ for which $2x^2 + px + 8$ is positive for all real $x$ .
Show that $x^2 + (k - 2)x - k = 0$ has two distinct real roots for all values of $k$ .
Find the set of values of $k$ for which $kx^2 - 4x + k = 0$ has no real roots.
The line $y = mx + 1$ is a tangent to the curve $y = x^2 + 2x + 5$ . Find the two possible values of $m$ and the corresponding points of contact.
Find the set of values of $k$ for which $(k - 1)x^2 + 2kx + (k + 2) = 0$ has two distinct real roots.
Find the set of values of $k$ for which the curve $y = kx^2 + kx + 1$ lies entirely above the $x$ -axis.

Answers

$(-5)^2 - 4(3)(4) = 25 - 48 = -23 < 0$ , so there are no real roots.
$36 - 4c = 0$ , so $c = 9$ . Then $x^2 - 6x + 9 = (x - 3)^2 = 0$ , root $x = 3$ .
$k^2 - 4(4)(9) = 0$ , so $k^2 = 144$ and $k = \pm 12$ .
$k^2 - 4(k + 3) > 0$ , i.e. $k^2 - 4k - 12 > 0$ , $(k - 6)(k + 2) > 0$ . So $k < -2$ or $k > 6$ .
$a = 2 > 0$ , so we need $p^2 - 64 < 0$ : $-8 < p < 8$ .
$b^2 - 4ac = (k - 2)^2 - 4(1)(-k) = k^2 - 4k + 4 + 4k = k^2 + 4$ . Since $k^2 \ge 0$ , this is at least $4$ , so always positive: two distinct real roots for every $k$ .
For $k \neq 0$ : $16 - 4k^2 < 0$ , so $k^2 > 4$ , giving $k < -2$ or $k > 2$ . For $k = 0$ the equation is $-4x = 0$ , which has the root $x = 0$ , so $k = 0$ is excluded (and it is not in the set anyway). Answer: $k < -2$ or $k > 2$ .
$x^2 + 2x + 5 = mx + 1$ gives $x^2 + (2 - m)x + 4 = 0$ . Tangent: $(2 - m)^2 - 16 = 0$ , so $2 - m = \pm 4$ , giving $m = -2$ or $m = 6$ .
- $m = 6$ : $x^2 - 4x + 4 = 0$ , $x = 2$ , $y = 13$ . Point $(2, 13)$ .
- $m = -2$ : $x^2 + 4x + 4 = 0$ , $x = -2$ , $y = 5$ . Point $(-2, 5)$ .
For the equation to be quadratic we need $k \neq 1$ . Discriminant: $(2k)^2 - 4(k - 1)(k + 2) = 4k^2 - 4\left(k^2 + k - 2\right) = 8 - 4k$ . Two distinct roots: $8 - 4k > 0$ , so $k < 2$ . When $k = 1$ the equation is $2x + 3 = 0$ , which has only one root. Answer: $k < 2$ , $k \neq 1$ .
If $k = 0$ , $y = 1$ , which is above the axis, so $k = 0$ works. If $k \neq 0$ we need $k > 0$ and $k^2 - 4k < 0$ , i.e. $k(k - 4) < 0$ , so $0 < k < 4$ . ( $k < 0$ gives a $\cap$ -shape, which must go below the axis.) Answer: $0 \le k < 4$ .

Where it comes from

Translating the wording

Practice questions

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