Download Quadratics Practice Answers

University of Southampton School of Education
24 week mathematics subject knowledge enhancement course
Developing Subject Knowledge – Quadratics answers
1. Using factorisation, solve each equation :
a) x² – 4x + 3 = 0
(x -1)(x- 3) = 0
x = 1 or 3
b) x(x – 1) = 42
x²- x -42 = 0
(x + 6)(x- 7) = 0
x = -6 or 7
d) 6 + 23x – 4x² = 0
4x² -23x- 6 = 0
(4x + 1)(x- 6) = 0
x = -¼ or 6
c) x² = 3x
x²-3x = 0
x(x- 3) = 0
x = 0 or 3
2. By completing the square, find the maximum or minimum value of y and the value of x for which
this occurs. State whether your value of y is a maximum or a minimum in each case.
a) y = x – 2x + 7
y = (x -1)² -1 + 7
y = (x -1)² + 6
y = 6 at x = 1, minimum
b) y = x² + 10x + 35
y = (x + 5)² -25 + 35
y = (x + 5)² + 10
y = 10 at x = -5, minimum
c) y = – x² + 4x + 4
y =- [x² -4x] + 4
y =- [(x -2)² - 4] + 4
y = -(x -2)² + 8
y = 8 at x = 2, maximum
d) y = 2x² + 8x + 5
y = 2x² + 8x + 5
y = 2(x² + 4x) + 5
y = 2((x + 2)² -4) + 5
y = 2(x + 2)² -8 + 5
y = 2(x + 2)² -3
y = -3 at x =- 2, minimum
3. Use the quadratic formula to solve each equation, giving your answers as simply as possible in
terms of surds where appropriate.
a) x² + 4x + 1 = 0
x = – 2 ±√3
b) – t² + 8t + 4 = 0
c) y² – 20y + 91 = 0
t = 4 ± 2√5
d) 5 – y – y² = 0
y = –½ (1 ±√21)
y = 7 or 13
4. By evaluating the discriminant, determine whether the roots of each equation are real and
distinct, real and equal or not real.
a) x² + 2x – 7 = 0
b² – 4ac = 32
real and distinct
b) x² + x + 3 = 0
c) x² + 14x + 49 = 0
b² – 4ac = – 11
not real
b² – 4ac = 0
real and equal
d) 5x² + 8x + 3 = 0
b² – 4ac = 4
real and distinct
5. Find the coordinates of the points of intersection of the given straight line and curve in each
case.
a) y = x + 2
y = x² – 4
(–2, 0) and (3, 5)
b) y = 4x + 11
y = x² + 3x – 1
(–3, –1) and (4, 27)
c) 2x – y = 5
x² + y² = 25
(0, –5) and (4, 3)
6. Giving your answers in terms of surds, solve the equation x4 – x² – 6 = 0
(x² + 2)(x² -3) = 0
x² = -2 [no solutions]
or
x² = 3
x = ±√3
7. Solve the equation x = 8√x – 15
let y = √x hence x = y²
x = 8√x – 15 becomes y² = 8y – 15
y² – 8y + 15 = 0
(y – 5)(y – 3) = 0
y = 5 or y = 3
x = 25 or x = 9
8. The diagram on the right shows the curve with equation
y = 2x² + px + q, where p and q are constants, which crosses
the x-axis at the points with coordinates (–2, 0) and (3, 0).
a) Show that p = –2 and find the value of q
Method 1
y = a(x + 2)(x – 3) (equation of graph must be of this form to cross x-axis at –2 and 3
= a(x² – x – 6)
= ax² – ax – 6a
ie coefficients of x² and x must be same magnitude but of opposite sign.
Since coefficient of x² is 2, coefficient of x (p) must also be –2
q = – 6a = –12
Method 2
When x = –2, y = 0 and when x = 3, y = 0
Substituting into : y = 2x² + px + q
gives 0 = 8 – 2p + q and 0 = 18 +3p + q
Solving these simultaneous equations gives p = –2 and q = –12
b) Find the coordinates of the vertex of the curve
Method 1
y = 2x² – 2x – 12
= 2[x² – x] – 12
= 2[(x – ½)² – ¼]– 12
= 2(x – ½)² – ½ – 12
= 2(x – ½)² – 12½
Vertex at (½,–12½)
Method 2
Equation of line of symmetry :
x = ½ (3 + -2) = ½
y = 2x² – 2x – 12 = – 12½