Question 1

LEVEL 6

Solve the following quadratic via factorisation:

2a^2 - 7a - 4 = 0

Select the correct answer from the options below.

A: a=-1, a=4

B: a=-\dfrac{1}{2}, a=4

C:\dfrac{1}{2}, a=-5

D: a=2, a=-4

 

CORRECT ANSWER:   B: a=-\dfrac{1}{2}, a=4

WORKED SOLUTION:

Firstly, multiply the coefficient of the a^2 term by the constant term, so 2 \times -4 = -8.

So, we want two numbers which multiply to make -8 and add to make -7.

The pair (1, -8) satisfies these criteria.

We should split up -7a accordingly into a – 8a, and rewrite our equation as

2a^2 + a - 8a - 4 = 0

Then, we factorise the first two terms and the last two terms. So, we get

a(2a + 1) - 4(2a + 1) = 0

Now, because (2a + 1) is a factor of both the terms on the left, we can factorise it out.

Note: at this point, if what’s inside both your brackets aren’t the same, then something has gone wrong with the method.

Thus, we get the full factorisation:

(2a + 1)(a - 4) = 0

Then, in order for the left-hand side to equal zero, we must have either

2a + 1 = 0 or a - 4 = 0.

Solving these two linear equations, we find that our two solutions are

a = -\frac{1}{2} and a = 4.

 

 

Question 2

LEVEL 6

Solve the following quadratic via factorisation:

3x^2 + x - 2 = 0

Select the correct answer from the options below.

A: x= 1.5, x=1

B:x= 2.5, x=-1

C:x= \dfrac{2}{3}, x=-1

D:x= \dfrac{3}{2}, x=-1

 

CORRECT ANSWER:  C: x= \dfrac{2}{3}, x=-1

WORKED SOLUTION:

Firstly, multiply the coefficient of the x^2 term by the constant term,

so 3 \times -2 = -6.

 

So, we want two numbers which multiply to make -6 and add to make 1. The pair (3, -2) works.

This pair then tells us we should split up x accordingly rewrite our equation as

3x^2 + 3x - 2x - 2 = 0

 

Then, we factorise the first two terms and the last two terms to get

3x(x + 1) - 2(x + 1) = 0

 

Now, because (x + 1) is a factor of both the terms on the left, we can factorise it out.

Note: at this point, if what’s inside both your brackets aren’t the same, then something has gone wrong.

(x + 1)(3x - 2) = 0

 

Then, in order for the left-hand side to equal zero, we must have either

x + 1 = 0 or 3x - 2 = 0.

 

Solving these two linear equations, we find that our two solutions are

x = -1 and x = \frac{2}{3}.

 

 

Question 3

LEVEL 6

Solve the following quadratic via factorisation:

6n^2 - 11n + 4 = 0

Select the correct answer from the options below.

A:x=\dfrac{4}{3}, x=-\dfrac{1}{2}

B:x=-\dfrac{4}{3}, x=\dfrac{1}{2}

C:x=4, x=\dfrac{1}{2}

D:x=\dfrac{4}{3}, x=\dfrac{1}{2}

 

CORRECT ANSWER:  D:x=\dfrac{4}{3}, x=\dfrac{1}{2}

WORKED SOLUTION:

Firstly, multiply the coefficient of the n^2 term by the constant term, so 6 \times 4 = 24.

So, we want two numbers which multiply to make 24 and add to make -11. The pair (-3, -8) work.

This pair then tells us how to split up our n term into -3n - 8n, and rewrite our equation as

6n^2 - 3n - 8n + 4 = 0

 

Then, we factorise the first two terms and the last two terms to get

3n(2n - 1) - 4(2n - 1) = 0

 

Now, because (2n - 1) is a factor of both the terms on the left, we can factorise it out.

Note: at this point, if what’s inside both your brackets aren’t the same, then something has gone wrong.

(2n - 1)(3n - 4) = 0

 

Then, in order for the left-hand side to equal zero,

we must have either 2n - 1 = 0 or 3n - 4 = 0.

 

Solving these two linear equations, we find that our two solutions are

n = \frac{1}{2} and n = \frac{4}{3}.

 

 

Question 4

LEVEL 6

Solve the following quadratic via factorisation:

9p^2 + 12p + 4=0

Select the correct answer from the options below.

A:p= -\dfrac{2}{3}

B:p= -\dfrac{2}{3}, p=\dfrac{2}{3}

C:p=\dfrac{2}{3}

D:p=\dfrac{3}{2}

 

CORRECT ANSWER:  A:p= -\dfrac{2}{3}

WORKED SOLUTION:

Firstly, multiply the coefficient of the p^2 term by the constant term, so 9 \times 4 = 36. So, we want two numbers which multiply to make 36 and add to make 12. The pair (6, 6) satisfies these criteria. This pair then tells us how to split up our p term, i.e. we should split up 12p accordingly into 6p + 6p, and rewrite our equation as

9p^2 + 6p + 6p + 4 = 0

Then, we treat the first two terms and the last two terms (on the left-hand side) as if they were separate expressions and factorise them as such. So, we get

3p(3p + 2) + 2(3p + 2) = 0

Now, because (3p + 2) is a factor of both the terms on the left, we can factorise it out. Note: at this point, if what’s inside both your brackets aren’t the same, then something has gone wrong with the method. Thus, we get the full factorisation:

(3p + 2)(3p + 2) = 0

The bracket is repeated, so in order for the left-hand side to equal zero, we must have that 3p + 2 = 0. Solving this equation, we get that our one solution is p = -\frac{2}{3}.

 

 

Question 5

LEVEL 6

Solve the following quadratic via factorisation:

16y^2 - 25=0

Select the correct answer from the options below.

A: y= -\dfrac{5}{4}, y= -\dfrac{5}{4}

B:y= \dfrac{5}{4}, y= \dfrac{5}{4}

C:y= \dfrac{5}{4}, y= -\dfrac{5}{4}

D:y= \dfrac{5}{4}

 

CORRECT ANSWER:  C:y= \dfrac{5}{4}, y= -\dfrac{5}{4}

WORKED SOLUTION:

This is a case of the difference of two squares.

By recognising this, and knowing that \sqrt{16} = 4 and \sqrt{25} = 5,

we can immediately factorise our equation and get

(4y - 5)(4y + 5) = 0

 

Then, in order for the left-hand side to equal zero,

we must have either 4y - 5 = 0 or 4y + 5 = 0.

 

Solving these two linear equations, we find that our two solutions are

y = \frac{5}{4} and y = -\frac{5}{4}.