Find k from the number of roots

In 5 of 20 recent exams
Paper 1 or 2
4 to 6 marks

How it is set in the exam

Mostly Paper 1 (no calculator), 4–6 marks; it also appears on Paper 2 (calculator), where the working is still algebra (Example 4.3). Typical (5 of 20 exam versions, 4 of them on Paper 1 (no calculator)). Sometimes a Section B "show that", or a follow-up about the signs of the roots.

Spot it in the exam

Exam phrases to spot (7)

“equal roots”
“two distinct real roots”
“no real roots”
“real roots”
“find the possible values of $k$ ”
“two distinct roots, both positive”
“roots of the same sign”

In disguise (10)

"the graph touches the x-axis" → $\Delta = 0$
"the equation has exactly one solution" → $\Delta = 0$ , and test the value of $k$ that makes the $x^2$ coefficient zero (it gives a linear equation, which also has one solution; see the discriminant)
" $f(x) > 0$ for all $x$ " or "the graph lies entirely above the x-axis" → $\Delta < 0$ and $a > 0$
"negative for all $x$ " → $\Delta < 0$ and $a < 0$
" $\geq 0$ for all $x$ " → $\Delta \le 0$ and $a > 0$
a log of a quadratic with $k$ in it, defined for every real $x$ (its domain is given as $x \in \mathbb{R}$ ) → the argument is positive for all $x$ : $\Delta < 0$ and $a > 0$ . With no unknown, finding where such a log is defined is Quadratic inequalities.
an equation in $e^{2x}$ and $e^x$ with $k$ in it that must have at least one solution → a quadratic in $u = e^x$ (Quadratic in disguise): $\Delta \ge 0$ , and a positive root $u$ , since $e^x > 0$ . If the axis $u = -\frac{b}{2a}$ is positive, the larger root is positive: say so, and the check is done. A $\ln k$ in the equation also needs $k > 0$ .
"both roots positive", "of the same sign", "one positive and one negative root" → $\Delta$ counts the roots but can't say where they are: add the sign of $f(0)$ and the side of the axis (step 5)
" $f(x) = k$ has one / two / no solutions" (or two positive ones), where $f$ is a fixed quadratic and $k$ is only in the constant term: that is Find the vertex (a horizontal line against a fixed curve: compare $k$ with the vertex's y-value), and $\Delta$ is an accepted alternative there. If the unknown is inside $f$ (in $a$ or $b$ ), rearrange to " $\ldots = 0$ " and use $\Delta$ here.
"a line is tangent to a curve" → see Two graphs

Worked example

Equal roots, with a trap

Paper 1 (no calculator)

Question

The equation $kx^2 - 2kx + 3 = 0$ has two equal real roots. Find the value of $k$ . [5]

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See it move: Change k until the roots are equal

k

1.5, no real roots

$k = 1.5$ : $a = 1.5$ , $b = -3$ , $c = 3$ , so $\Delta = (-3)^{2} - 4(1.5)(3) = -9$ : no real roots: the graph stays off the axis.
"Equal roots" means $\Delta = 0$ : an equation in $k$ , so specific values rather than a range. State the condition before you use it: it earns its own mark.
$\Delta = (-2k)^{2} - 4(k)(3) = 4k^{2} - 12k = 4k(k - 3) = 0$ , so $k = 0$ or $k = 3$ . The question says "the value", singular: one of them goes.
$k = 0$ makes the $x^{2}$ coefficient zero: the equation becomes $3 = 0$ , which is not a quadratic and has no roots. Reject it.
$k = 3$ : $3x^{2} - 6x + 3 = 3(x - 1)^{2}$ , the repeated root $x = 1$ . For $k < 0$ or $k > 3$ the roots are distinct; between them there are none.

Steps

Rearrange to $ax^2 + bx + c = 0$ and write down $a$ , $b$ , $c$ . They will contain $k$ .
Translate the wording into a condition on $\Delta$ (table in the discriminant) and state it ("two distinct real roots, so $\Delta > 0$ "): the stated condition earns its own mark (usually ). Then write $\Delta = b^2 - 4ac$ with the values substituted. $\Delta$ seen only inside the quadratic formula often scores nothing.
Solve: $\Delta = 0$ is an equation in $k$ ; $\Delta > 0$ or $\Delta < 0$ is an inequality in $k$ (sketch it).
Find the value of $k$ that makes the $x^2$ coefficient zero in the original equation or inequality, and test it against the wording. Reject it for "equal roots" and "two distinct real roots". It can be valid for "exactly one solution" or "positive for all $x$ ". Apply any condition on $a$ (for example $a > 0$ for "positive for all $x$ ").
5
Signs of the roots
(for $a > 0$ ; if $a < 0$ , multiply the equation by $-1$ first, which doesn't change the roots). Sketch a U-shape crossing the x-axis twice:
- one positive and one negative root: $f(0) < 0$ (the graph crosses the y-axis below the x-axis; $\Delta > 0$ then comes free)
- both roots the same sign: $\Delta > 0$ and $f(0) > 0$
- both positive: also the axis to the right of the y-axis, $-\frac{b}{2a} > 0$ (both negative: $-\frac{b}{2a} < 0$ )
Then combine all the conditions on a number line.

Common mistakes

Wrong condition
"real roots" is $\geq 0$ , not $> 0$ .
$\Delta$ only inside the quadratic formula
writing $x = \frac{-k \pm \sqrt{k^2 - 4(k+3)}}{2}$ often earns nothing for $\Delta$ . Write $\Delta = \ldots$ on its own line with the values, and state the condition ( $\Delta > 0$ ): that statement is usually a mark of its own.
Using only $\Delta$ for "both positive"
$\Delta > 0$ says there are two roots, not where they are. Add $f(0) > 0$ and the side of the axis.

More mistakes (7)

Solving $\Delta > 0$ as if it were an equation
and giving $k = -2, 6$ instead of a range.
Ignoring $a$
not testing the value that makes $a = 0$ , or forgetting to require $a > 0$ in "positive for all $x$ " questions.
Leaving a rejected value in the answer
rejected values must go. In Example 4.1, " $k = 0$ or $k = 3$ " as the final answer scores once $k = 0$ has been rejected.
Leaving $k$ out of $c$
in $x^2 + kx + k + 3$ , $c = k + 3$ , not $3$ (and in Example 4.3, $c = p - 2$ ).
Sign slips in $b^2$
$(-2k)^2 = 4k^2$ , not $-4k^2$ . Put brackets around $b$ before squaring.
Multiplying or dividing an inequality by a negative without flipping
$-4p^2 + 8p + 9 < 0$ becomes $4p^2 - 8p - 9 > 0$ , not $< 0$ ; and $16 - 4m^2 < 0$ gives $m^2 > 4$ , not $m^2 < 4$ .
Square-rooting an inequality
$m^2 > 4$ is not $m > \pm 2$ . It is $m < -2$ or $m > 2$ . Sketch it.

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