IGCSE · Practice questions · The discriminant: roots, lines and curves

Cross, touch or miss.

Six original Cambridge-style questions on the discriminant: the number of roots, values of \(k\) for equal, distinct or no real roots, and whether a line meets, touches or misses a curve, with points of contact.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

One quadratic, one discriminant, one decision.

Line and curve? Set the two expressions for \(y\) equal first.
Rearrange to \(ax^{2} + bx + c = 0\) and read \(a\), \(b\), \(c\) with their signs.
Translate the words: two distinct points \(> 0\); equal roots or tangent \(= 0\); no real roots or does not meet \(< 0\); meets \(\ge 0\).
Solve for the constant; an inequality in \(k\) needs its critical values and the correct one-piece or two-piece answer.
Point of contact: put the value back, solve the repeated root, then find \(y\) from the line.

Recall

[2 marks]

Use the discriminant to determine the number of real roots of the equation \(x^{2} + 7 = 3x\).

Rearrange: \(x^{2} - 3x + 7 = 0\), so \(a = 1,\ b = -3,\ c = 7\). ✓
\(b^{2} - 4ac = 9 - 28 = -19 < 0\), so there are no real roots. ✓

Calculation

[3 marks]

Find the value of \(k\) for which the equation \(kx^{2} - 6x + 3 = 0\) has equal roots, and find the root.

Equal roots: \((-6)^{2} - 4(k)(3) = 0\), so \(36 - 12k = 0\), giving \(k = 3\). ✓
\(3x^{2} - 6x + 3 = 3(x - 1)^{2} = 0\). ✓
The root is \(x = 1\). ✓

Calculation

[4 marks]

Find the set of values of \(k\) for which the equation \(x^{2} + 2kx + 3k + 4 = 0\) has no real roots.

No real roots: \((2k)^{2} - 4(1)(3k + 4) < 0\). ✓
\(4k^{2} - 12k - 16 < 0\); divide by 4: \(k^{2} - 3k - 4 < 0\). ✓
\((k - 4)(k + 1) < 0\); critical values \(-1\) and \(4\). ✓
Below the axis between the roots: \(-1 < k < 4\). ✓

Analysis

[4 marks]

The line \(y = x + k\) is a tangent to the curve \(y = 2x^{2} - 3x + 4\).

(a) Find the value of \(k\).

(b) Find the coordinates of the point of contact.

(a) \(2x^{2} - 3x + 4 = x + k\), so \(2x^{2} - 4x + (4 - k) = 0\). ✓
Tangent: \(16 - 4(2)(4 - k) = 0\), so \(16 - 32 + 8k = 0\), giving \(k = 2\). ✓
(b) \(2x^{2} - 4x + 2 = 2(x - 1)^{2} = 0\), so \(x = 1\). ✓
\(y = 1 + 2 = 3\); the point of contact is \((1,\ 3)\). ✓

Analysis

[6 marks]

The line \(y = mx + 2\) and the curve \(y = x^{2} + 3x + 6\) are given.

(a) Find the set of values of \(m\) for which the line meets the curve at two distinct points.

(b) For each value of \(m\) for which the line is a tangent to the curve, find the coordinates of the point of contact.

(a) \(x^{2} + 3x + 6 = mx + 2\), so \(x^{2} + (3 - m)x + 4 = 0\). ✓
Two distinct points: \((3 - m)^{2} - 16 > 0\), so \((m - 3)^{2} > 16\). ✓
\(m - 3 > 4\) or \(m - 3 < -4\), so \(m > 7\) or \(m < -1\). ✓
(b) Tangent when \(m = 7\) or \(m = -1\). ✓
\(m = 7\): \(x^{2} - 4x + 4 = (x - 2)^{2} = 0\), so \(x = 2\), \(y = 7(2) + 2 = 16\); point \((2,\ 16)\). ✓
\(m = -1\): \(x^{2} + 4x + 4 = (x + 2)^{2} = 0\), so \(x = -2\), \(y = 2 + 2 = 4\); point \((-2,\ 4)\). ✓

Proof

[4 marks]

Show that the line \(y = k(x - 1)\) meets the curve \(y = x^{2} - 1\) for every value of \(k\), and find the value of \(k\) for which the line is a tangent to the curve.

\(x^{2} - 1 = kx - k\), so \(x^{2} - kx + (k - 1) = 0\). ✓
\(b^{2} - 4ac = k^{2} - 4(k - 1) = k^{2} - 4k + 4 = (k - 2)^{2}\). ✓
A square is never negative, so \(b^{2} - 4ac \ge 0\) for every \(k\): the line always meets the curve. ✓
Tangent when \((k - 2)^{2} = 0\), that is \(k = 2\) (the line \(y = 2x - 2\) touches the curve at \((1,\ 0)\)). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.