Quadratic Equations Long Questions (Question 11 – 13)

Posted on May 25, 2020 by Myhometuition

Question 11:
Solve the following quadratic equation:

$\frac{5 x + 3 x^{2}}{1 - 2 x} = 6$

Solution:

$\begin{array}{l} \frac{5 x + 3 x^{2}}{1 - 2 x} = 6 \\ 5 x + 3 x^{2} = 6 - 12 x \\ 3 x^{2} + 5 x + 12 x - 6 = 0 \\ 3 x^{2} + 17 x - 6 = 0 \\ (3 x - 1) (x + 6) = 0 \\ 3 x - 1 = 0 or x + 6 = 0 \\ 3 x = 1 x = - 6 \\ x = \frac{1}{3} x = - 6 \end{array}$

Question 12:

Diagram above shows a rectangle ABCD.

(a) Express the area of ABCD in terms of n.

(b) Given the area of ABCD is 60 cm², find the length of AB.

Solution:

(a)

Area of ABCD

= (n + 7) × n

= (n²+ 7n) cm²

(b)

Given the area of ABCD = 60

n²+ 7n = 60

n²+ 7n – 60 = 0

(n – 5) (n + 12) = 0

n = 5 or n = – 12 (not accepted)

When n = 5,

Length of AB = 5 + 7 = 12 cm

Question 13 (4 marks):
Solve the following quadratic equation:

$- \frac{2}{3 x - 5} = \frac{x}{3 x - 1}$

Solution:

$\begin{array}{l} - \frac{2}{3 x - 5} = \frac{x}{3 x - 1} \\ - 2 (3 x - 1) = x (3 x - 5) \\ - 6 x + 2 = 3 x^{2} - 5 x \\ 3 x^{2} - 5 x + 6 x - 2 = 0 \\ 3 x^{2} + x - 2 = 0 \\ (3 x - 2) (x + 1) = 0 \\ 3 x - 2 = 0 or x + 1 = 0 \\ x = \frac{2}{3} or x = - 1 \end{array}$

SPM Practice (Paper 1)

Posted on May 22, 2020 by Myhometuition

Question 14:

$Show that 6 x - 6 - 2 k x^{2} = x^{2} has no real roots if k > \frac{1}{4} .$

Solution:

Question 15:
The quadratic equation

$x^{2} + p x + q = 0$ has roots –2 and 6. Find
(a) the value of p and of q,
(b) the range of values of r for which the equation

$x^{2} + p x + q = r$ has no real roots.

Solution:

SPM Practice (Paper 1)

Posted on May 22, 2020 by Myhometuition

Question 9:
The roots of the equation

$6 x^{2} + h x + 1 = 0 are α and β, whereas 3 α and 3 β$ are the roots of the equation

$2 x^{2} - x + k = 0$ . Find the value of h and k.

Solution:

$\begin{array}{l} 6 x^{2} + h x + 1 = 0 \\ a = 6, b = h, c = 1 \\ Roots = α, β \\ sor : \\ α + β = - \frac{b}{a} \\ α + β = - \frac{h}{6} ......... (1) \\ p o r : \\ α β = \frac{c}{a} \\ α β = \frac{1}{6} ......... (2) \\ 2 x^{2} - x + k = 0 \\ a = 2, b = - 1, c = k \\ Roots = 3 α, 3 β \\ sor : \\ 3 α + 3 β = - \frac{b}{a} \\ 3 (α + β) = - \frac{(- 1)}{2} \\ α + β = \frac{1}{6} ......... (3) \\ p o r : \\ 3 α (3 β) = \frac{c}{a} \\ 9 α β = \frac{k}{2} \\ k = 18 α β ......... (4) \\ Substitute (3) into (1) \\ α + β = - \frac{h}{6} \\ \frac{1}{6} = - \frac{h}{6} \\ h = - 1 \\ Substitute (2) into (4) \\ k = 18 α β \\ k = 18 (\frac{1}{6}) \\ k = 3 \end{array}$

Question 10:
Find the range of values of p for which the equation

$2 x^{2} + 5 x + 3 - p = 0$ has two real distinct roots.

Solution:

SPM Practice (Paper 1)

Posted on May 22, 2020 by Myhometuition

Question 3:
Solve the following quadratic equations by using quadratic formula. Give your answer in four significant figures.

$\begin{array}{l} (a) (x + 1) (x - 5) = 15 \\ (b) \frac{x^{2} + 3 x - 2}{x^{2} - x - 1} = 3 \end{array}$

Solution:

Question 4:
If the roots of

$2 x^{2} + 4 x - 1 = 0 are α and β,$ find the equations whose roots are

$\begin{array}{l} (a) α^{2}, β^{2} \\ (b) α - β, β - α \end{array}$

Solution:

Question 5:
Write and simplify the equation whose roots are double the roots of

$3 x^{2} - 5 x - 1 = 0$ , without solving the given equation.

Solution:

Quadratic Equations, SPM Practice (Paper 2)

Posted on May 15, 2020 by Myhometuition

2.10.3 Quadratic Equations, SPM Practice (Paper 2)

Question 5:

Given 3t and (t – 7) are the roots of the quadratic equation 4x² – 4x + m = 0 where m is a constant.

(a) Find the values of t and m.

(b) Hence, form the quadratic equation with roots 4t and 2t + 6.

Solutions:

(a)

Given 3t and (t – 7) are the roots of the quadratic equation 4x² – 4x + m = 0

a = 4, b = – 4, c = m

Sum of roots =

$- \frac{b}{a}$

3t + (t– 7) =

$- \frac{- 4}{4}$

3t + t– 7 = 1

4t = 8

t = 2

Product of roots =

$\frac{c}{a}$

3t (t– 7) =

$\frac{m}{4}$

4 [3(2) (2 – 7)] = m ← (substitute t = 2)

4 [3(2) (2 – 7)] = m

4 (–30) = m

m = –120

(b)

t = 2

4t = 4(2) = 8

2t + 6 = 2(2) + 6 = 10

Sum of roots = 8 + 10 = 18

Product of roots = 8(10) = 80

Using the formula, x²– (sum of roots)x + product of roots = 0

Thus, the quadratic equation is,

x² – 18x + 80 = 0

Quadratic Equations, SPM Practice (Paper 2)

Posted on April 18, 2020 by user

Question 4:
It is given α and β are the roots of the quadratic equation x (x – 3) = 2k – 4, where k is a constant.

$\begin{array}{l} (a) Find the range of values of k if α \neq β . \\ (b) Given \frac{α}{2} and \frac{β}{2} are the roots of another quadratic equation \\ 2 x^{2} + t x - 4 = 0, where t is a constant, find the value of t and of k . \end{array}$

Solution:

$\begin{array}{l} (a) \\ x (x - 3) = 2 k - 4 \\ x^{2} - 3 x + 4 - 2 k = 0 \\ a = 1, b = - 3, c = 4 - 2 k \\ b^{2} - 4 a c > 0 \\ {(- 3)}^{2} - 4 (1) (4 - 2 k) > 0 \\ 9 - 16 + 8 k > 0 \\ 8 k > 7 \\ k > \frac{7}{8} \end{array}$

$\begin{array}{l} (b) \\ From the equation x^{2} - 3 x + 4 - 2 k = 0, \\ α + β = - \frac{b}{a} \\ = - \frac{- 3}{1} \\ = 3............. (1) \\ α β = \frac{c}{a} \\ = \frac{4 - 2 k}{1} \\ = 4 - 2 k ............. (2) \\ From the equation 2 x^{2} + t x - 4 = 0, \\ \frac{α}{2} + \frac{β}{2} = - \frac{t}{2} \\ α + β = - t ............. (3) \\ \frac{α}{2} \times \frac{β}{2} = - \frac{4}{2} \\ α β = - 8............. (4) \\ Substitute (1) = (3), \\ 3 = - t \\ t = - 3 \\ Substitute (2) = (4), \\ 4 - 2 k = - 8 \\ 4 + 8 = 2 k \\ k = 6 \end{array}$

Quadratic Equations, SPM Practice (Paper 2)

Posted on April 18, 2020 by user

2.10.2 Quadratic Equations, SPM Practice (Paper 2)

Question 3:

If α and β are the roots of the quadratic equation 3x² + 2x– 5 = 0, form the quadratic equations that have the following roots.

$\begin{array}{l} (a) \frac{2}{α} and \frac{2}{β} \\ (b) (α + \frac{2}{β}) and (β + \frac{2}{α}) \end{array}$

Solution:

3x² + 2x – 5 = 0

a = 3, b = 2, c = –5

The roots are α and β.

$\begin{array}{l} α + β = - \frac{b}{a} = - \frac{2}{3} \\ α β = \frac{c}{a} = \frac{- 5}{3} \end{array}$

(a)

$\begin{array}{l} The new roots are \frac{2}{α} and \frac{2}{β} . \\ Sum of new roots \\ = \frac{2}{α} + \frac{2}{β} = \frac{2 β + 2 α}{α β} \\ = \frac{2 (α + β)}{α β} = \frac{2 (- \frac{2}{3})}{- \frac{5}{3}} = \frac{4}{5} \end{array}$

$\begin{array}{l} Product of new roots \\ = (\frac{2}{α}) (\frac{2}{β}) = \frac{4}{α β} \\ = \frac{4}{- \frac{5}{3}} = - \frac{12}{5} \end{array}$

Using the formula, x²– (sum of roots)x + product of roots = 0

The new quadratic equation is

$x^{2} - (\frac{4}{5}) x + (- \frac{12}{5}) = 0$

5x² – 4x– 12 = 0

(b)

$\begin{array}{l} The new roots are (α + \frac{2}{β}) and (β + \frac{2}{α}) . \\ Sum of new roots \\ = (α + \frac{2}{β}) + (β + \frac{2}{α}) \end{array}$

$\begin{array}{l} = α + β + (\frac{2}{α} + \frac{2}{β}) = α + β + \frac{2 α + 2 β}{α β} \\ = α + β + \frac{2 (α + β)}{α β} \\ = \frac{4}{5} + \frac{2 (\frac{4}{5})}{- \frac{12}{5}} \\ = \frac{4}{5} - \frac{2}{3} = \frac{2}{15} \end{array}$

$\begin{array}{l} Product of new roots \\ = (α + \frac{2}{β}) (β + \frac{2}{α}) \\ = α β + 2 + 2 + \frac{4}{α β} \end{array}$

$\begin{array}{l} = - \frac{12}{5} + 4 + \frac{4}{- \frac{12}{5}} \\ = - \frac{12}{5} + 4 - \frac{5}{3} = - \frac{1}{15} \end{array}$

The new quadratic equation is

$x^{2} - (\frac{2}{15}) x + (- \frac{1}{15}) = 0$

15x² – 2x– 1 = 0

Quadratic Equations, SPM Practice (Paper 2)

Posted on April 18, 2020 by user

Question 2:

Given α and β are two roots of the quadratic equation (2x + 5)(x + 1) + p = 0 where αβ = 3 and p is a constant.

Find the value p, α and of β.

Solutions:

(2x + 5)(x + 1) + p = 0

2x² + 2x + 5x + 5 + p = 0

2x² + 7x + 5 + p = 0

*Compare with, x²– (sum of roots)x + product of roots = 0

$x^{2} + \frac{7}{2} x + \frac{5 + p}{2} = 0 \leftarrow \begin{array}{l} divide both \\ sides with 2 \end{array}$

Product of roots, αβ = 3

$\frac{5 + p}{2} = 3$

5 + p = 6

p = 1

Sum of roots =

$- \frac{7}{2}$

$\begin{array}{l} α + β = - \frac{7}{2} \to (1) \\ and α β = 3 \to (2) \\ from (2), β = \frac{3}{α} \to (3) \\ Substitute (3) into (1), \\ α + \frac{3}{α} = - \frac{7}{2} \end{array}$

2α²+ 6 = –7α ← (multiply both sides with 2α)

2α²+ 7α + 6 = 0

(2α + 3)(α + 2) = 0

2α + 3 = 0 or α + 2 = 0

α=− 3 2 α = –2

$\begin{array}{l} Substitute α = - \frac{3}{2} into (3), \\ β = \frac{3}{- \frac{3}{2}} = 3 (- \frac{2}{3}) = - 2 \end{array}$

Substitute α = –2 into (3),

$\begin{array}{l} β = - \frac{3}{2} \\ ∴ p = 1, and \\ when α = - \frac{3}{2}, β = - 2 and α = - 2, β = - \frac{3}{2} . \end{array}$

Quadratic Equations, SPM Practice (Paper 2)

Posted on April 18, 2020 by user

2.10.1 Quadratic Equations, SPM Practice (Paper 2)

Question 1:

(a) Find the values of k if the equation (1 – k) x²– 2(k + 5)x + k + 4 = 0 has real and equal roots.

Hence, find the roots of the equation based on the values of k obtained.

(b) Given the curve y = 5 + 4x – x²has tangen equation in the form y = px + 9. Calculate the possible values of p.

Solutions:

(a)

For equal roots,

b²– 4ac = 0

[–2(k + 5)]² – 4(1 – k)( k + 4) = 0

4(k + 5)²– 4(1 – k)( k + 4) = 0

4(k² + 10k + 25) – 4(4 – 3k – k²) = 0

4k² + 40k + 100 – 16 + 12k + 4k² = 0

8k² + 52k + 84 = 0

2k² + 13k + 21 = 0

(2k + 7) (k + 3) = 0

$k = - \frac{7}{2}, - 3$

$k = - \frac{7}{2}$ , the equation is

$\begin{array}{l} (1 + \frac{7}{2}) x^{2} - 2 (- \frac{7}{2} + 5) x - \frac{7}{2} + 4 = 0 \\ \frac{9}{2} x^{2} - 3 x + \frac{1}{2} = 0 \end{array}$

9x² – 6x + 1 = 0

(3x – 1) (3x – 1) = 0

x = ⅓

If k = –3, the equation is

(1 + 3)x²– 2(–3 + 5)x – 3 + 4 = 0

4x² – 4x + 1 = 0

(2x – 1) (2x – 1) = 0

x = ½

(b)

y = 5 + 4x – x²----- (1)

y = px + 9 ---------- (2)

(1) = (2), 5 + 4x – x²= px + 9

x² + px – 4x + 9 – 5 = 0

x² + (p – 4)x + 4 = 0

Tangen equation only has one intersection point with equal roots.

b²– 4ac = 0

(p – 4)² – 4(1)(4) = 0

p² – 8p + 16 – 16 = 0

p² – 8p = 0

p (p – 8) = 0

Therefore, p = 0 and p = 8.

2.2a Solving Quadratic Equations – Factorisation

Posted on April 18, 2020 by user

2.4.1 Solving Quadratic Equations – Factorisation

1. If a quadratic equation can be factorised into a product of two factors such that

(x – p)(x – q) = 0

Hence

x – p = 0 or x – q = 0

x = p or x = q

p and q are the roots of the equation.

Notes
1.The equation must be written in general form ax² + bx+ c = 0 before factorisation.

2. This method can only be used if the quadratic expression can be factorised completely.

Example 1:
Find the roots of the quadratic equations
(a) x (2x − 8) = 0
(b) x² −16x = 0
(c) 3x² − 75x = 0
(d) 5x² − 100x = 25x

Solution:

(a)

x (2x − 8) = 0

x = 0 or 2x − 8 = 0

2x − 8 = 0

2x = 8

x= 4

x = 0 or x = 4

(b)

x² −16x = 0

x (x − 16) = 0

x = 0 or x − 16 = 0

x = 0 or x = 16

(c)

3x² − 75x = 0

3x (x − 25) = 0

3x = 0 or x − 25 = 0

x = 0 or x = 25

(d)

5x² − 100x = 25x

5x² − 100x − 25x = 0
5x² − 125x = 0

x (5x − 125) = 0

x = 0 or 5x − 125 = 0

5x = 125

x = 25

x = 0 or x = 25

Example 2:

Solve the following quadratic equations
(a) x² − 4x – 5 = 0
(b) 1 − 5x + 2x² = 4

Solution:

(a)

x² − 4x – 5 = 0
(x – 5) (x + 1) = 0

x – 5 = 0 or x + 1 = 0

x = 5 or x = –1

(b)

1 − 5x + 2x² = 4

2x² − 5x + 1 – 4 = 0

2x² − 5x – 3 = 0

(2x + 1) (x – 3) = 0

2x + 1= 0 or x – 3 = 0

2x = –1 or x = 3

x = –½ or x = 3