# Algebra Higher Revision Card Answers

## HA1 – Rules of Indices

QUESTION: Write the expression below in the form $\sqrt[a]{x^b}$ where $a$ and $b$ are integers to be found.

$\dfrac{\left(x^{\dfrac{2}{3}}\right)^6}{x^{\dfrac{5}{3}}\times x^{-\dfrac{1}{3}}}$

ANSWER: First up, the numerator. The term has a power on top of another power, so they should be multiplied to get

$\left(x^{\dfrac{2}{3}}\right)^6=x^{\dfrac{2}{3}\times 6}=x^{\dfrac{12}{3}}$

At this point, we could simplify the fractional power to just be 4, but it will actually make the last step easier if we leave it as a fraction.

Now, the denominator. The terms are being multiplied, so we add the powers. Note: we are going to add a positive number to a negative number in exactly the same we always add a positive number to a negative number. Doing so, we get

$x^{\dfrac{5}{3}}\times x^{-\dfrac{1}{3}}=x^{\dfrac{5}{3}+\left(-\dfrac{1}{3}\right)}=x^{\dfrac{4}{3}}$

Next, we can treat the whole fraction as a division (meaning we will subtract the powers). So, after having simplified the top and bottom we get

$\dfrac{x^{\dfrac{12}{3}}}{x^{\dfrac{4}{3}}}=x^{\dfrac{12}{3}-\dfrac{4}{3}}=x^{\dfrac{8}{3}}$

According to the rules of fractional powers, this can be written like

$x^{\dfrac{8}{3}}=\sqrt[3]{x^8}$

We have found that $a=3, b=8$, so we’re done.

## HA2 – Negative Indices

QUESTION: Evaluate the expression below. Leave your answer in its simplest form.

$12^{-2}\times 36^{\dfrac{1}{2}}$

ANSWER: Firstly, simplify $12^{-2}$. We need to take the reciprocal of 12, and then make the power positive. Doing so, we get

$12^{-2}=\dfrac{1}{12^2}=\dfrac{1}{144}$

Next, $36^{\frac{1}{2}}$. A fractional power becomes a root, so we get

$36^{\dfrac{1}{2}}=\sqrt{36}=6$

Therefore, the original expression is

$\dfrac{1}{144}\times6=\dfrac{6}{144}$

This simplifies to

$\dfrac{3}{72}$

This simplifies further to

$\dfrac{1}{24}$

## HA3 – Rearranging Formulae

QUESTION: Rearrange the following formula to make $t$ the subject.

$v=\dfrac{1}{2}at^2$

ANSWER: Firstly, we will multiply both sides by 2 to get

$2v=at^2$

Then, dividing both sides by $a$ gives us

$\dfrac{2v}{a}=t^2$

We are almost there but not quite – we need to get rid of the power of 2. To do that, we must square root both sides to get

$\sqrt{\dfrac{2v}{a}}=t$

Flipped around (this step is not necessary, but is good practice), this looks like

$t=\sqrt{\dfrac{2v}{a}}$

## HA4 – Forming & Solving Equations

QUESTION: At the cinema, one large drink costs £4 and one adult ticket costs £$x$. Laline buys 2 adult tickets and one large drink.

a) Write an expression, in terms of $x$, for the total cost of Laline’s cinema trip.

b) The total cost of Laline’s trip was £26. Use your answer to part a) to form an equation, and

use that equation to determine the value of $x$.

ANSWER: a) Laline bought two tickets, so that’s two lots of £$x$, and one large drink, which is one lot of £4. Adding these together, we get the expression

$2x+4$

b) The total cost of her trip was £26, and we know the expression above also represents the total cost of her trip, so we can set them equal to each other. Doing so, we get

$2x+4=26$

We can now solve this equation. Subtracting 4 from both sides, we get

$2x=26-4=\pounds 22$

Then, dividing both sides by 2 gives us the answer

$x=22\div2=\pounds 11$.

## HA5 – Algebraic Fractions

Question: Fully simplify the algebraic fraction below.

$\dfrac{x^2 + 3x}{4x + 12}$

ANSWER: Factorising the numerator and denominator of our algebraic fraction, we see that

$\dfrac{x^2 + 3x}{4x + 12} = \dfrac{x(x + 3)}{4(x + 3)}$

We see that top and bottom have a common factor $(x + 3)$, so it can cancel. This leaves us with

$\dfrac{x}{4}$

This can not be further simplified, and so we are done.

## HA6 – Surds 1: Basics

QUESTION: Write $\sqrt{72}$ in simplified surd form.

ANSWER: We want to find a square number that goes into 72. Here, we will go for 36. So, we can write $\sqrt{72}$ like

$\sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2}$

We know $\sqrt{36}=6$, so we get

$\sqrt{36}\times\sqrt{2}=6\sqrt{2}$

There are no square numbers bigger than 1 that go into 2, so we are done.

## HA7 – Surds 2: Expanding Brackets & Rationalising the Denominator

QUESTION: Rationalise the denominator of the fraction below. Write your answer in its simplest form.

$\dfrac{8}{3\sqrt{6}}$

ANSWER: We need to multiply top and bottom by the surd that’s on the bottom: $\sqrt{6}$. Doing so, the numerator becomes

$8\times\sqrt{6}=8\sqrt{6}$

And the denominator becomes

$3\sqrt{6}\times\sqrt{6}=3\sqrt{36}=3\times6=18$

Therefore, the fraction becomes

$\dfrac{8\sqrt{6}}{18}$

We can cancel out a factor of 2 from the top and bottom, meaning the simplified fraction is

$\dfrac{4\sqrt{6}}{9}$

## HA8 – Expanding Brackets

QUESTION:

1) Expand $pqr(5pr+5r^5-25pqr)$.

2) Expand and simplify $(y-6)(y-2)$.

1) We must multiply the term on the outside, $pqr$, by all the terms on the inside. To do this, we must use the law of indices that says: when you multiply terms, their powers are added. So, multiplying the terms out, we get

\begin{aligned}pqr\left(5pr+5r^5-25pqr\right)&=\left(pqr\times 5pr\right)+\left(pqr\times 5r^5\right)+\left(pqr\times \left(-25pqr\right)\right) \\ &=5p^{2}qr^2+5pqr^6-25p^{2}q^{2}r^{2}\end{aligned}

2) We will apply FOIL, drawing red lines as we go, and then collect like terms once the expansion is done. So, we get

QUESTION: Factorise $k^2-7k+10$.

ANSWER: We want two numbers which multiply to make 10 and add to make -7. In order for them to multiply to make a positive number but add to make a negative one, we need them to both be negative. So, considering factors of 10,

$10=(-1)\times(-10)=(-2)\times(-5)$

We see that $(-2)\times(-5)=10$ and $-2+(-5)=-7$, so the two numbers we want are -2 and -5. Therefore, the quadratic factorises to

$k^2-7k+10=(k-2)(k-5)$

## HA10 – Solving Quadratics by Factorisation

QUESTION: Solve $z^2-2z-24=0$.

ANSWER: We want two numbers which multiply to make -24 and add to make -2. So, considering factors of -24,

$-24=1\times(-24)=2\times(-12)=3\times(-8)=4\times(-6)$

We see that $4\times(-6)=-24$ and $4+(-6)=-2$, so the two numbers we want are 4 and -6. Therefore, the equation becomes

$(z-6)(z+4)=0$

Therefore, the two solutions are

$z=6\,\,\,\,\text{and}\,\,\,\,z=-4$

## HA11 – The Quadratic Formula

QUESTION: Solve $2t^2-5t+2=0$

ANSWER: In this case, $a=2, b=-5,$ and $c=2$. Putting these into the formula, we get

$t=\dfrac{-(-5)\pm\sqrt{(-5)^2 -4\times 2 \times 2}}{2\times 2}=\dfrac{5\pm\sqrt{25-16}}{4}$

Putting this into a calculator, we get the solutions to be

$t=2,\,\,\,\,\,\text{and}\,\,\,\,\,t=\dfrac{1}{2}$

## HA12 – Completing the Square

QUESTION: Use completing the square to write  $z^2-8z+14$ in the form $(x+a)^2 +b$

ANSWER: We want to write this quadratic in the form $(z+a)^2+b$. Firstly, $a$ is half the coefficient of $z$ so $a=-8\div2=-4$. Then, $b$ is 14 takeaway the square of $a$, so we get

\begin{aligned}z^2 -8z+14&=(z-4)^2+14-(-4)^2 \\ &=(z-4)^2+14-16 \\ &=(z-4)^2 -2\end{aligned}

Therefore, our equation becomes:

$(z-4)^2-2$

As we can see this is in the form $(x+a)^2 +b$, so $(z-4)^2-2$ is our final answer.

Extra:

Now are have the equation in the form $(x+a)^2 +b$, we can actually solve for $x$ by doing the following.

Adding 2 to both sides and then square rooting, we get

\begin{aligned}(z-4)^2&=2\\z-4&=\pm\sqrt{2}\end{aligned}

Then, adding 4 to both sides, we get the solution to be

$z=4\pm\sqrt{2}$

In other words, the 2 solutions are

$z=4+\sqrt{2},\,\,\,\,\text{and}\,\,\,\,z=4-\sqrt{2}$

## HA13 – Linear Sequences & the nth Term

QUESTION: Find the nth term of the sequence below.

$1,\,\,\,5,\,\,\,9,\,\,\,13,\,\,\,17$

ANSWER: The formula must take the form $an+b$. To find $a$, find the common difference between the terms and confirm that they are the same.

So, the formula must be $4n+b$. So, we now write out the sequence given by $4n$ (i.e., the 4 times table):

$4,\,\,\,8,\,\,\,12,\,\,\,16,\,\,\,20$

Each of these terms is 3 bigger than their respective terms in the original sequence, so we must subtract 3 from them all. Therefore, the nth term is

$4n-3$

## HA14 – The nth Term of a Quadratic Sequence

QUESTION: Find the nth term of the quadratic sequence below.

$-1,\,\,\,4,\,\,\,15,\,\,\,32,\,\,\,55$

ANSWER: Firstly, we must find the second differences of the sequence.

The second difference is 6, so we get that $a=6\div2=3$. Now, we want to rewrite the sequence, with the values generated by $3n^2$ underneath it. Then, we will subtract the second row from the first to find a sequence of differences.

The sequence, in this case, is

$-4,\,\,-8,\,\,-12,\,\,-16,\,\,-20$

The common difference of this sequence is -4, so the the nth term for this sequence must begin with $-4n$. In fact, $-4n$ is precisely the nth formula for this sequence – no need to add/subtract anything. Combining this with $3n^2$, we get the nth term formula of the quadratic sequence to be

$3n^2-4n$

## HA15 – Inequalities

QUESTION: Solve the inequality below and plot your answer on a number line.

$2-4z\leq z-18$

ANSWER: We will rearrange to make $z$ the subject. So, adding $4z$ to both sides, we get

$2\leq 5z-18$

Then, adding 18 to both sides, we get

$20\leq 5z$

Finally, dividing both sides by 5 gives us the solution

$4\leq z$

So now we need to draw a closed circle at 4 on the number line (the inequality is inclusive), and draw an arrow going to the right, away from the circle. This looks like

QUESTION: Find the solutions to the inequality below.

$x^2+5x+6\leq0$

ANSWER: We want to find the two roots of this quadratic in order to draw a sketch of it. So, noticing that $2\times3=6$ and $2+3=5$, we get

$x^2+5x+6=(x+2)(x+3)$

Therefore, the two roots of this quadratic are $x=-2\text{ and }x=-3$. Therefore, the sketch of the graph looks like

Since we are looking for when $x^2+5x+6$ is less than or equal to zero, we want to see when the graph goes below the x-axis. We can see that that happens between -2 and -3, and so the solution to our inequality is

$-3\leq x\leq-2$

## HA17 – Iterative Methods

QUESTION:

a) Show that $x^3+2x=4$ can be rearranged to $x=\dfrac{4}{x^2+2}$.

b) Using the iterative formula below, with $x_0=1$, find the value of $x_1, x_2,$ and $x_3$, all to 3sf.

$x_{n+1}=\dfrac{4}{x_{n}^{2}+2}$

a) There is an $x^2$ in the end result but no such term in the initial equation. So, our first step is going to be factorising $x$ out of the left-hand side of the equation:

$x(x^2+2)=4$

Then, dividing both sides by $(x^2+2)$, we get

$x=\dfrac{4}{x^2+2}$

b) We need to sub in $x_0$ to the formula to get $x_1$, then sub in $x_1$ to get $x_2$, and so on. NOTE: we will round each decimal to 3sf in our answers, but in the calculation, we will use the full decimal expansions given by the calculator. So, we get

$x_1=\dfrac{4}{x_{0}^{2}+2}=\dfrac{4}{(1)^2+2}=\dfrac{4}{3}=1.3333...$

$x_2=\dfrac{4}{x_{1}^{2}+2}=\dfrac{4}{(1.3333...)^2+2}=\dfrac{18}{17}=1.058...$

$x_3=\dfrac{4}{x_{2}^{2}+2}=\dfrac{4}{(1.058...)^2+2}=1.281...$

Therefore, the rounded answers are $x_1=1.33, x_2=1.06, x_3=1.28$.

## HA18 – Linear Simultaneous Equations

QUESTION: Solve the following simultaneous equations.

\begin{aligned}2a-4b&=-1 \\ 4a+6b&=12\end{aligned}

ANSWER: To make the coefficients of $a$ the same, we will multiply the first equation by 2. Doing so, we get

$4a-8b=-2$

Then, we will subtract the second equation from this new equation we just obtained. This looks like

Thus, we get the equation $-14b=-14$. If we divide both sides by $-14$, we get

$b=\dfrac{-14}{-14}=1$

Then, substituting $b=1$ into our very first equation, we get

$2a-4=-1$

$2a=3$

Then, divide both sides by 2 to get

$a=\dfrac{3}{2}$

We have found both $a$ and $b$, so we’re done.

NOTE: if your first step was to divide the second equation given in the question by 2, then that is perfectly valid. If you got the right answer, then that method (or any other way of multiply/dividing equations to make the elimination step work) is worth full marks.

## HA19 – Quadratic Simultaneous Equations

QUESTION: Solve the following simultaneous equations.

$x^2+y^2=2$

$y=2x-1$

ANSWER: We want to eliminate a variable by using substitution. So, we’re going to replace the $y$ in the first equation with $2x-1$, as the second equation tells us that $y$ is equal to $2x-1$. Doing this, we get

$x^2+(2x-1)^2=2$

Expanding the brackets and collecting like terms, we get

$x^2+4x^2-2x-2x+1=2$

$5x^2-4x+1=2$

$5x^2-4x-1=0$

This is a quadratic we can solve. This can be done using whichever method you prefer, here we will use the quadratic formula. Doing so, we get

$x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4\times5\times(-1)}}{2\times 5}$

$x=\dfrac{4\pm\sqrt{16+20}}{10}$

Putting it all into a calculator, we get the solutions to be

$x=1,\,\,\,\,x=-\dfrac{1}{5}$

Then, we find the solutions for $y$ by substituting these values into the original equation. Doing so, we get

$x=1\,\,\text{gives}\,\,y=2(1)-1=1$

$x=-\dfrac{1}{5}\,\,\text{gives}\,\,y=2\left(-\frac{1}{5}\right)-1=-\dfrac{7}{5}$

Therefore, the two solution pairs are

$x=1,\,y=1\,\,\,\,\text{and}\,\,\,\,x=-\dfrac{1}{5},\,y=-\dfrac{7}{5}$

## HA20 – Inverse and Composite Functions

QUESTION: Let $f(x)=4x^3 and g(x)=\dfrac{x-1}{5}$.

a) i) Find $f^{-1}(x)$.

ii) Evaluate $f^{-1}(32)$.

b) Rey claims that for a fixed value of $x$, $gf(x)$ is always bigger than $fg(x)$. Find a counterexample to show she is wrong.

ANSWER: a)i) We want to rearrange $y=4x^3$ to make $x$ the subject. So, dividing both sides by 4 we get

$\dfrac{y}{4}=x^3$

Then, cube rooting both sides, we get

$x=\sqrt[3]{\dfrac{y}{4}}$

Finally, switching any $y$ with an $x$, we get that the inverse function is

$f^{-1}(x)=\sqrt[3]{\dfrac{x}{4}}$

ii) We want to sub 32 into $f^{-1}(x)$. Doing so, we get

$f^{-1}(32)=\sqrt[3]{\dfrac{32}{4}}=\sqrt[3]{8}=2$

b) So, firstly we get

\begin{aligned}fg(0)=f\left(\frac{0-1}{5}\right)&=f\left(\frac{-1}{5}\right) \\ &=4\left(\frac{-1}{5}\right)^3 \\ &=-0.032\end{aligned}

And then, we get

\begin{aligned}gf(0)=g\left(4(0)^3\right)&=g(0) \\ &=\dfrac{0-1}{5} \\ &=-0.2\end{aligned}

$-0.032>-0.2$, therefore in this case, $fg(0)>gf(0)$, so it must not always be true that $gf(x)>fg(x)$.

## HA21 - Proof

QUESTION: Prove that $(2m+3)^2-(2m-3)^2$ is always a multiple of 6.

ANSWER: To do this, we need to expand the brackets and collect like terms. So, the first bracket:

\begin{aligned}(2m+3)^2&=4m^2+6m+6m+9 \\ &=4m^2+12m+9\end{aligned}

Next, the second bracket:

\begin{aligned}(2m-3)^2&=4m^2-6m-6m+9 \\ &=4m^2-12m+9\end{aligned}

Therefore, subtracting one from another, we get

\begin{aligned}(2m+3)^2-(2m-3)^2&=4m^2+12m+9 - (4m^2-12m+9) \\ &=12m -(-12m)=24m\end{aligned}

So, the expression simplifies to $24m$. 24 is a multiple of 6, so we can write

$24m=6\times4m$

Since $4m$ must be a whole number, we can conclude that $24m$, and therefore the original expression, must be a multiple of 6.