14. Solving quadratics by factorising (a=1)

Question 1

LEVEL 4

Solve the following quadratic via factorisation:

x^2 + 3x - 4 = 0

Select the correct answer from the options below.

B: $x=1$ and $x=-4$

C: $(x + 1)(x - 4)$

D: $(x + 4)(x + 1)$

CORRECT ANSWER: A: $x=1$ and $x=-4$

WORKED SOLUTION:

We are looking for two numbers which multiply together to make $-4$ and add to make $3$ .

The factors of $-4$ are

$1$ , $-4$

$-1$ , $4$

$2$ , $-2$

and $-1 +4= 3$ so the correct pair is $-1$ and $4$ .

(x -1)(x + 4) = 0

When $x=1$ or $x=-4$ the equation equals $0$ as shown

$x=1$ and $x=-4$

Question 2

LEVEL 4

Solve the following quadratic via factorisation:

b^2 - 14b + 48 = 0

Select the correct answer from the options below.

A: $b=-6$ and $b=-8$

B: $b=-6$ and $b=8$

D: $b=6$ and $b=-8$

CORRECT ANSWER: C: $b=6$ and $b=8$

WORKED SOLUTION:

We are looking for two numbers which multiply together to make $48$ and add to make $-14$ .

The factors of $48$ are

$8$ , $6$

$-8$ , $-6$

$2$ , $24$

$-2$ , $-24$

$4$ , $12$

$-4$ , $-12$

and $-6 -8= -14$ so the correct pair is $-6$ and $-8$ .

(b -6)(b -8) = 0

When $b=6$ or $b=8$ the equation equals $0$ as shown

$b=6$ and $b=8$

Question 3

LEVEL 4

Solve the following quadratic via factorisation:

x^2 - 49=0

Select the correct answer from the options below.

A: $x=-7$ and $x=1$

B: $x=-7$

C: $x=7$

CORRECT ANSWER: D: $x=-7$ and $x=7$

WORKED SOLUTION:

This is a case of the difference of two squares.

Recognising this, and observing that $\sqrt{49} = 7$ ,

we can immediately factorise this quadratic equation as such:

(x + 7)(x - 7) = 0

$x = -7$ and $x = 7$

Question 4

LEVEL 4

Solve the following quadratic via factorisation:

a^2 - 10a + 25 = 0

Select the correct answer from the options below.

A: $a=-5$ , $a=5$

B: $a=-5$

D: $a=10$

CORRECT ANSWER: C: $a=5$

WORKED SOLUTION:

We are looking for two numbers which multiply together to make $25$ and add to make $-10$ .

The only viable factorisation is $-5 \times -5$ . So, our factorised equation is

(a - 5)(a - 5) = 0

As both the brackets are the same, it must be

a - 5 = 0

So the answer is $a = 5$

Question 5

LEVEL 4

Solve the following quadratic via factorisation:

m^2 - 16m + 28 = 0

Select the correct answer from the options below.

B: $m = -14$ , $m = -2$

C: $m = -14$ , $m = 2$

D: $m = 14$ , $m = -2$

CORRECT ANSWER: A: $m = 14$ , $m = 2$

WORKED SOLUTION:

We are looking for two numbers which multiply together to make $28$ and add to make $-16$ .

We see that some possible factorisations of $28$ are

28 = -1 \times -28 = -2 \times -14 = -4 \times -7

Noticing that $-2 + (-14) = -16$ ,

we can see this must be the correct pairing. So, our factorised quadratic equation is

(m - 2)(m - 14) = 0

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

$m - 2 = 0$ or $m - 14 = 0$ ,

thus our two solutions are $m = 2$ , and $m = 14$ .

Question 6

LEVEL 4

Solve the following quadratic via factorisation:

x^2+7x+6=0

Select the correct answer from the options below.

B: $x=1$ and $x=-6$

C: $x=-1$ and $x=7$

D: $x=1$ and $x=7$

CORRECT ANSWER: A: $x=-1$ and $x=-6$

WORKED SOLUTION:

So, we must factorise this quadratic. Observing that $6\times1=6$ and $6+1=7$ , the quadratic factorises to

$x^2+7x+6=(x+1)(x+6)$

Therefore, we can rewrite the quadratic equation in the question to be

$(x+1)(x+6)=0$

This now gives us two very simple equations to solve:

$x+1=0\,\text{ and }\,x+6=0$ .

To solve the first equation, subtract 1 from both sides to get $x=-1$ . To solve the second equation, subtract $6$ from both sides to get $x=-6$ . Thus, the two solutions are $x=-1$ and $x=-6$ .

Note: You may have realised that the two solutions are just the numbers in the factorisation, $(x+1)(x+6)$ , with opposite signs.