Questions on quadratic functions

Question 1

Express x² + 8x + 3 in the form (x + p)² + q, and state the minimum point of the curve y = x² + 8x + 3.

Accepted Answer

Half the 8, square it, correct for it: (x + 4)² − 16 + 3 = (x + 4)² − 13. The square is smallest when its bracket is zero, so the minimum point is (−4, −13). M1 for halving and squaring the 8, A1 for the completed square, B1 for the minimum point. Reading the minimum as (4, −13) is the sign slip to watch, since the bracket vanishes at x = −p.

Question 2

Use the discriminant to determine how many real roots 2x² − 3x + 5 = 0 has.

Accepted Answer

b² − 4ac = 9 − 40 = −31, which is negative, so there are no real roots. M1 for the discriminant, A1 for the conclusion. The parabola opens upward and its minimum sits above the axis, so it never touches down.

Question 3

The same quadratic is written in three forms: y = x² − 2x − 8, y = (x − 4)(x + 2) and y = (x − 1)² − 9. Write down the roots, the coordinates of the vertex and the y-intercept of the curve, stating which form gives each directly.

Accepted Answer

The factorised form gives the roots, where a bracket is zero: x = 4 and x = −2. B1. The completed square gives the vertex, where the square is zero: (1, −9), a minimum since the square is added. B1. The expanded form gives the y-intercept as its constant term, (0, −8). B1. A vertex read as (−1, −9) has the sign inside the bracket backwards, since (x − 1)² vanishes at x = 1, and the three answers are consistent, the vertex sitting halfway between the roots.

Question 4

Show that x² + 6x + 10 is positive for every real value of x.

Accepted Answer

Complete the square: x² + 6x + 10 = (x + 3)² + 1. M1. A square is never negative, so the expression is at least 1 for every x, and in particular is positive. A1 for the conclusion drawn from the square. The discriminant offers a second route: 36 − 40 = −4 is negative, so the upward parabola never meets the axis and stays above it.

Question 5

Express 3x² − 12x + 5 in the form a(x + p)² + q, and state the minimum point.

Accepted Answer

Take the 3 out of the x terms first. 3(x² − 4x) + 5 = 3[(x − 2)² − 4] + 5, and distributing gives 3(x − 2)² − 12 + 5 = 3(x − 2)² − 7. The minimum point is (2, −7). M1 for taking the 3 out, A1 for the bracket, A1 for −7, B1 for the minimum point. That correction term must be multiplied by the 3 on the way back out, and leaving it as −4 + 5 = 1 omits that factor of 3.

Question 6

Find the values of k for which x² + kx + 25 = 0 has a repeated root.

Accepted Answer

A repeated root means the discriminant is zero: k² − 100 = 0, so k = 10 or k = −10. M1 for setting the discriminant to zero, A1 for both values. Two different parabolas, each just touching the axis, one touching to the left of the origin and one to the right.

Question 7

The equation kx² + 4x + 1 = 0 has two distinct real roots. Find the set of possible values of k.

Accepted Answer

Two distinct roots need 16 − 4k > 0, so k < 4. But k = 0 has to be excluded as well, because the equation is then linear and has a single root rather than two. So k < 4 and k ≠ 0. M1 for a positive discriminant, A1 for k < 4, B1 for excluding k = 0.

Question 8

Solve 9^x − 10 × 3^x + 9 = 0.

Accepted Answer

Since 9^x = (3^x)², substitute u = 3^x to get u² − 10u + 9 = 0, which factorises as (u − 1)(u − 9) = 0. Then 3^x = 1 gives x = 0, and 3^x = 9 gives x = 2. M1 for the substitution, A1 for the factorised quadratic, A1 for x = 0, A1 for x = 2.

Question 9

Sketch the curve y = 2(x − 3)² − 8, showing the coordinates of its vertex and of every point where it meets the coordinate axes.

Accepted Answer

The completed square gives the vertex at once: the bracket is zero at x = 3, so the minimum point is (3, −8), and the positive 2 says the parabola opens upwards. B1. At x = 0, y = 2(9) − 8 = 10, so the y-intercept is (0, 10). B1. For the roots set y = 0: 2(x − 3)² = 8, so (x − 3)² = 4 and x − 3 = ±2, giving x = 1 and x = 5. M1 for solving from the completed square with both signs, A1 for (1, 0) and (5, 0). The sketch is a U through those four points, symmetric about x = 3.

Question 10

A ball is thrown upwards and its height above the ground after t seconds is modelled by h = 20t − 5t² metres. By completing the square, find the greatest height reached and the time at which it occurs, and find when the ball lands.

Accepted Answer

Take out the −5 from the t terms: h = −5(t² − 4t) = −5[(t − 2)² − 4] = 20 − 5(t − 2)². M1 for completing the square with the −5 taken out. The subtracted square is smallest, at zero, when t = 2, so the greatest height is 20 m at t = 2 s. A1. The ball lands when h = 0: 5t(4 − t) = 0, so t = 0, the throw, or t = 4. M1 for setting h = 0 and factorising, A1 for t = 4 s. The negative leading coefficient is why the vertex is a maximum, and it multiplies the correction term on the way out: leaving the −4 inside as −4 instead of +20 gives a peak of −4 m, which a thrown ball cannot manage. The landing time is twice the time to the peak, as the symmetry of the parabola says it should be.

Question 11

Find the set of values of k for which the equation x² + (k − 2)x + k + 1 = 0 has no real roots.

Accepted Answer

No real roots means a negative discriminant: (k − 2)² − 4(k + 1) < 0. M1 for the discriminant condition with the correct coefficients. Expanding, k² − 4k + 4 − 4k − 4 = k² − 8k, so k(k − 8) < 0. A1 for the quadratic in k. That is an upward parabola in k, below the axis strictly between its roots. M1 for solving the inequality from the roots 0 and 8. So 0 < k < 8. A1. The unknown appears in b and in c at once, so the condition is a quadratic inequality in k rather than a linear one, and a check at k = 4 confirms the interval: x² + 2x + 5 has discriminant 4 − 20 < 0. At k = 0 the equation is x² − 2x + 1 = (x − 1)², a repeated root, which is why the ends of the interval are excluded.

Question 12

Given that the equation (k + 1)x² + 4kx + 9 = 0 has a repeated root, find the two possible values of k.

Accepted Answer

Set the discriminant to zero: (4k)² − 4(k + 1)(9) = 0, so 16k² − 36k − 36 = 0. Dividing by 4 gives 4k² − 9k − 9 = 0, which factorises as (k − 3)(4k + 3) = 0, so k = 3 or k = −3/4. M1 for setting the discriminant to zero, A1 for the quadratic in k, dM1 for factorising it, A1 for k = 3, A1 for k = −3/4. Both are legitimate: k = 3 turns the equation into 4x² + 12x + 9 = (2x + 3)² = 0, and k = −3/4 into (x − 6)²/4 = 0. Only k = −1 would need excluding, and neither answer is −1.

Question 13

Show that for every real value of k the equation x² − 2kx + k² − 4 = 0 has two distinct real roots, and find the roots in terms of k.

Accepted Answer

The discriminant is (−2k)² − 4(k² − 4) = 4k² − 4k² + 16 = 16. M1 for the discriminant with all three coefficients in place, A1 for 16. It is positive whatever k is, since k has cancelled out entirely, so there are always two distinct real roots. B1 for the conclusion, with the reason that 16 does not depend on k. The roots follow from the formula or from the square: x² − 2kx + k² = (x − k)², so the equation is (x − k)² = 4 and x − k = ±2. M1 for completing the square or applying the formula with √16 = 4. So x = k + 2 and x = k − 2. A1. The two roots are always 4 apart, which is the square root of the discriminant over the leading coefficient, and the curve is y = x² − 4 translated k to the right.

Question 14

By completing the square, show that the solutions of ax² + bx + c = 0, where a ≠ 0, are given by x = (−b ± √(b² − 4ac))/(2a). State the condition on a, b and c for the equation to have real solutions, and explain where it arises in your derivation.

Accepted Answer

Divide by a, which is allowed because a ≠ 0: x² + (b/a)x + c/a = 0. M1. Complete the square on the first two terms: (x + b/(2a))² − b²/(4a²) + c/a = 0. M1 for halving b/a and correcting by its square. Rearranging and putting the right side over 4a²: (x + b/(2a))² = b²/(4a²) − c/a = (b² − 4ac)/(4a²). A1 for that line. Take square roots, keeping both signs: x + b/(2a) = ±√(b² − 4ac)/(2a), since √(4a²) = 2|a| and the ± absorbs the sign of a. M1 for the square root with ±. Subtracting b/(2a) gives x = (−b ± √(b² − 4ac))/(2a). A1. Real solutions need a real square root, so the condition is b² − 4ac ≥ 0; it arises at the square-root step, because the left side is a square and cannot equal a negative number, and the right side has the positive denominator 4a², so its sign is the sign of b² − 4ac. B1 for the condition and its origin. Equality gives one repeated root, −b/(2a), the x-coordinate of the vertex.

Question 15

The equation x² + kx + k = 0 has two real roots that differ by 3. Find the possible values of k, giving your answers in exact form.

Accepted Answer

By the formula the roots are (−k ± √(k² − 4k))/2. M1 for the roots in terms of k. Their difference is the gap between the ± versions, √(k² − 4k), since the −k/2 parts cancel. A1 for the difference. Setting it equal to 3 and squaring: k² − 4k = 9, so k² − 4k − 9 = 0. M1 for forming the equation in k. The formula, or completing the square as (k − 2)² = 13, gives k = 2 + √13 or k = 2 − √13. A1 A1 for the two values. Both are admissible: the roots are real exactly when k² − 4k ≥ 0, and for either value k² − 4k = 9, which is positive, so the two roots exist and are 3 apart in each case. Squaring the difference was safe because a difference of 3 and a square root are both non-negative, so no false solution was created. As a check, k = 2 + √13 ≈ 5.61 gives roots of about −1.30 and −4.30, which are 3 apart.