Question 1

LEVEL 8

Solve the inequality x^2 \leq 81.

Select the correct answer from the list below:

A: x < -9 \text{ and } x > 9

B: x \leq -9 \text{ and } x \geq 9

C: - 9\leq x \leq 9

D: - 9 < x < 9

 

CORRECT ANSWER:  C: - 9\leq x \leq 9

WORKED SOLUTION:

Square root both sides of the equation. We get 2 square roots, positive and negative.

The range of values lie between and including -9 and 9, so

- 9\leq x \leq 9

Question 2

LEVEL 8

Solve the inequality 4 x^2 > 144.

Select the correct answer from the list below:

A: x < -6 \text{ and } x > 6

B: x \leq -6 \text{ and } x \geq 6

C: - 6\leq x \leq 6

D: - 6 < x < 6

 

CORRECT ANSWER:  A: x < -6 \text{ and } x > 6

WORKED SOLUTION:

Square root both sides of the equation. We get 2 square roots, positive and negative.

The range of values lie below -6 and above 6, so

x < -6 \text{ and } x > 6

Question 3

LEVEL 8

Solve the inequality x^2+7x+12<0.

Select the correct answer from the list below:

A: x<3\text{ and }x>4

B: 3<x<4

C: -4<x<-3

D: x< -4\text{ and } x>-3

CORRECT ANSWER:   C: -4<x<-3

WORKED SOLUTION:

To solve this inequality, we need to factorise the quadratic. Observing that 3\times 4=12 and 3+4=7, we get:

x^2+7x+12=(x+3)(x+4)<0

If we treat this like an equality to start off with, we can use this to plot the quadratic as y=(x+3)(x+4). Note that the graph will cross the x-axis at x=-4 and x=-3.

Now, we don’t want all these values we want all the ones that are (x+3)(x+4)<0. Because we plotted this as y=(x+3)(x+4), we can say that y<0 and so only want the bit below the y-axis.

So, we can see that we only want the values of x that are between -4 and -3. Therefore, the solution to the inequality is:

-4<x<-3

Question 4

LEVEL 8

Solve the inequality x^2+3x-10\geq0.

 

Select the correct answer from the list below:

A: x\leq-5 \text{ and }x\geq2

B: -5<x<2

C: -5\leq x\leq2 

D: x\leq-2 \text{ and }x\geq5

 

CORRECT ANSWER:  A: x\leq-5 \text{ and }x\geq2

WORKED SOLUTION:

To solve this inequality, we need to factorise the quadratic. Observing that (-2)\times5=10 and -2+5=3, we get:

x^2+7x+12=(x-2)(x+5)\geq0

If we treat this like an equality to start off with, we can use this to plot the quadratic as y=(x-2)(x+5). Note that the graph will cross the x-axis at x=-5 and x=2.

Now, we don’t want all these values we want all the ones that are (x-2)(x+5)\geq0. Because we plotted this as y=(x-2)(x+5), we can say that y\geq0 and so only want the bits above the y-axis.

So, we can see that we only want the values of x that are less than -5 and bigger than 2. Therefore, the solution to the inequality is:

x\leq-5 \text{ and }x\geq2

Question 5

LEVEL 8

Solve the inequality x^2-4x>0.

Select the correct answer from the list below:

A: -2<x<2

B: x<-2 \text{ and }x>2

C: x<0 \text{ and }x>4

D: 0<x<4

 

CORRECT ANSWER:  C: x<0 \text{ and }x>4

WORKED SOLUTION:

To solve this inequality, we need to factorise the quadratic. Factorising here is pretty simple, as we can just pull out an x.

x^2-4x=x(x-4)>0

If we treat this like an equality to start off with, we can use this to plot the quadratic as y=x(x-4). Note that the graph will cross the x-axis at x=0 and x=4.

Now, we don’t want all these values we want all the ones that are x(x-4)>0. Because we plotted this as y=x(x-4), we can say that y>0 and so only want the bits above the y-axis.

So, we can see that we only want the values of x that are less than 0 and bigger than 4. Therefore, the solution to the inequality is:

x<0 \text{ and }x>4