When forming an expression for a surface area, what is easy to leave out?

The faces that are hidden in the diagram, since only some of a solid's faces are visible from one viewpoint. Count the faces first, note which of them are identical, then write an area expression for each one.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsAlgebra Toolkit

Algebra Toolkit (AQA GCSE Further Maths): Flashcards

Exam code: 8365

1/56

0Still learning

Know0

FrontExpanding & Factorising
True or False?
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2}$

Was this flashcard helpful?

Cards in this collection (56)

True or False?
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2}$
False.
A squared bracket must be written out in full as $(x + y + z) (x + y + z)$ and then expanded.
Squaring each term separately misses every cross term, such as the $2 x y$ .
How do you expand $(2 x + 5) (3 x - 4 y + 5)$ ?
Split it into two single-bracket expansions, one for each term in the first bracket:
$2 x (3 x - 4 y + 5) + 5 (3 x - 4 y + 5) = 6 x^{2} - 8 x y + 25 x - 20 y + 25$
FOIL does not work here, because it only accounts for four products rather than six.
Complete the factorisation by filling in the highest common factor:
$6 x^{4} y^{7} - 8 x^{9} y^{5} =______(3 y^{2} - 4 x^{5})$
The completed factorisation is:
$6 x^{4} y^{7} - 8 x^{9} y^{5} = 2 x^{4} y^{5} (3 y^{2} - 4 x^{5})$
Take the smallest power of each variable that appears, since that is the most every term is guaranteed to contain.
Factorise fully $(x + 1) (2 x + 3) + (x + 1) (x - 2)$ .
Take out the common bracket $(x + 1)$ , then simplify what is left inside:
$(x + 1) [(2 x + 3) + (x - 2)] = (x + 1) (3 x + 1)$
A highest common factor can be a whole bracket, not just a number or a variable.
How do you expand three brackets multiplied together?
Expand any two of them first and collect like terms, then multiply that result by the remaining bracket.
The second stage is usually of the form $(x + a) (x^{2} + b x + c)$ , which produces six terms before simplifying.
Complete the factorisation by grouping:
$a b + 3 b + 2 a + 6 = b (a + 3) +______(a + 3) = (a + 3) (______)$
The completed factorisation is:
$a b + 3 b + 2 a + 6 = b (a + 3) + 2 (a + 3) = (a + 3) (b + 2)$
Once both pairs produce the same bracket, that bracket comes out as a factor. The pairing is not always the first two terms and the last two; look for which terms share a factor.
Expand $(2 x - 3) (x + 4) (3 x - 1)$ .
Expanding the first two brackets gives $2 x^{2} + 5 x - 12$ .
Multiplying that by $(3 x - 1)$ and collecting like terms gives:
$6 x^{3} + 13 x^{2} - 41 x + 12$
Define quadratic expression.
A quadratic expression has the form $a x^{2} + b x + c$ , where $a \neq 0$ .
Any higher power of $x$ , such as an $x^{3}$ term, means the expression is not quadratic.
Complete the rule for factorising $x^{2} + b x + c$ by inspection:
Find two numbers that multiply to $______$ and add to $______$ .
The completed rule is:
Find two numbers that multiply to $c$ and add to $b$ .
For $x^{2} - 2 x - 8$ that pair is $- 4$ and $+ 2$ , giving $(x + 2) (x - 4)$ .
When factorising $a x^{2} + b x + c$ with $a \neq 1$ , what must the two numbers multiply to?
The two numbers must multiply to $a c$ , not to $c$ , while still adding to $b$ .
For $4 x^{2} - 25 x - 21$ that means multiplying to $4 \times (- 21) = - 84$ and adding to $- 25$ , which gives $- 28$ and $+ 3$ .
Factorise $6 x^{2} - 7 x - 3$ .
Split the middle term using $+ 2 x$ and $- 9 x$ , which multiply to $- 18$ and add to $- 7$ :
$6 x^{2} + 2 x - 9 x - 3 = 2 x (3 x + 1) - 3 (3 x + 1) = (3 x + 1) (2 x - 3)$
The shared bracket $(3 x + 1)$ then comes out as a factor.
Complete the difference of two squares:
$a^{2} - b^{2} = (a + b) (______)$
The completed identity is:
$a^{2} - b^{2} = (a + b) (a - b)$
The two brackets can be written the other way round, but the $a$ must come first in each, because it is $a^{2} - b^{2}$ and not $b^{2} - a^{2}$ .
True or False?
$x^{2} + 4$ cannot be factorised.
True.
The difference of two squares needs a subtraction, so a sum of two squares does not factorise.
$x^{2} - 4$ factorises to $(x + 2) (x - 2)$ , but $x^{2} + 4$ has no such pair of brackets.
How do you factorise $3 x^{2} + 13 x y - 10 y^{2}$ ?
Factorise the easier quadratic $3 x^{2} + 13 x - 10$ first, which gives $(3 x - 2) (x + 5)$ .
Then put a $y$ on the last term of each bracket, giving $(3 x - 2 y) (x + 5 y)$ .
Factorise fully $20 x^{3} - 45 x$ .
Take out the common factor $5 x$ first, which reveals a difference of two squares:
$20 x^{3} - 45 x = 5 x (4 x^{2} - 9) = 5 x (2 x + 3) (2 x - 3)$
Here $4 x^{2} = {(2 x)}^{2}$ and $9 = 3^{2}$ , so the bracket is a difference of two squares even though it does not look like one at first.
Factorise ${(7 x + 3)}^{2} - {(3 x - 2)}^{2}$ .
Use $a^{2} - b^{2} = (a + b) (a - b)$ with $a = 7 x + 3$ and $b = 3 x - 2$ :
$((7 x + 3) + (3 x - 2)) ((7 x + 3) - (3 x - 2)) = (10 x + 1) (4 x + 5)$
The two squared quantities can be whole brackets rather than single terms.
How can you tell whether a quadratic expression will factorise?
Work out the discriminant, $b^{2} - 4 a c$ : if it is a square number, the expression factorises.
Solving the quadratic on a calculator works too, since whole-number or fraction solutions mean it factorises and surd solutions mean it does not.
Which two factorising methods apply to a quadratic with only two terms?
Take out the highest common factor, as in $x^{2} - 7 x = x (x - 7)$ .
If instead it is one square subtracted from another, such as $x^{2} - 9$ , use the difference of two squares.
Why should you check for a common factor before factorising $3 x^{2} + 15 x + 18$ ?
Because 3 divides every term, so taking it out gives $3 (x^{2} + 5 x + 6)$ and the bracket now starts with $x^{2}$ .
A quadratic starting with $x^{2}$ is much easier to factorise, giving $3 (x + 2) (x + 3)$ .
True or False?
In $3 x^{2} - 5 x - 2$ , the 3 can be taken out as a common factor.
False.
3 divides $3 x^{2}$ but it does not divide $- 5 x$ or $- 2$ , so there is no common factor to take out.
This one has to be factorised as a harder quadratic, giving $(3 x + 1) (x - 2)$ .
The discriminant of $3 x^{2} + 7 x - 6$ is 121. Does it factorise?
Yes: $121 = 11^{2}$ , which is a square number, so the expression factorises.
If the discriminant were not a square number, the solutions would involve surds and no factorisation with whole numbers would be possible.
Factorise fully $- 8 x^{2} + 100 x - 48$ .
Take out $- 4$ first, which gives $- 4 (2 x^{2} - 25 x + 12)$ .
Factorising the bracket as a harder quadratic gives:
$- 8 x^{2} + 100 x - 48 = - 4 (2 x - 1) (x - 12)$
Complete the rule for completing the square:
$x^{2} + b x = {(x +______)}^{2} - {(______)}^{2}$
The completed rule is:
$x^{2} + b x = {(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2}$
Half of $b$ goes inside the bracket and the same quantity is then squared and subtracted, so $x^{2} + 8 x = {(x + 4)}^{2} - 16$ .
Write $x^{2} + 10 x + 9$ in completed square form.
Replace $x^{2} + 10 x$ with ${(x + 5)}^{2} - 25$ , then add the 9:
$x^{2} + 10 x + 9 = {(x + 5)}^{2} - 16$
The $- 25$ and the $+ 9$ combine, which is why the number at the end is not the original 9.
True or False?
$x^{2} - 20 x = {(x - 10)}^{2} + 100$
False.
The squared number is always subtracted, so $x^{2} - 20 x = {(x - 10)}^{2} - 100$ .
A negative $b$ changes the sign inside the bracket, not the sign in front of the number at the end.
How do you complete the square when there is a number in front of $x^{2}$ ?
Factor that number out of the $x^{2}$ and $x$ terms only, using square brackets, and leave the constant sitting outside.
Complete the square inside the brackets, then multiply back out and combine the numbers at the very end.
Write $4 x^{2} + 16 x + 5$ in the form $a {(x + p)}^{2} + q$ .
Take the 4 out of the first two terms only, then complete the square inside:
$4 [x^{2} + 4 x] + 5 = 4 [{(x + 2)}^{2} - 4] + 5 = 4 {(x + 2)}^{2} - 11$
The $+ 5$ takes no part at all until the last step, when $- 16$ and $+ 5$ combine.
Complete the coordinates of the turning point of $y = a {(x + p)}^{2} + q$ :
$(______,______)$
The turning point is at:
$(- p, q)$
Note the negative on the $x$ -coordinate. It is a minimum when $a > 0$ and a maximum when $a < 0$ , but the coordinates are the same either way.
Find the turning point of $y = x^{2} + 6 x - 11$ .
Completing the square gives $y = {(x + 3)}^{2} - 20$ .
The turning point is therefore at $(- 3, - 20)$ .
How does completing the square show that $y = x^{2} + 6 x - 3$ has a minimum value of $- 12$ ?
Completing the square gives $y = {(x + 3)}^{2} - 12$ .
A squared bracket can never be negative, so its smallest possible value is 0, which makes the smallest possible value of $y$ equal to $- 12$ .
A quadratic has a minimum turning point at $(2, - 5)$ . Write down a possible equation for it.
One possibility is $y = {(x - 2)}^{2} - 5$ .
Reading $(- p, q)$ backwards gives $p = - 2$ and $q = - 5$ , and any positive number in front of the bracket would also work.
True or False?
To solve $x (x - 4) = 0$ you can divide both sides by $x$ .
False.
Dividing by $x$ throws away the solution $x = 0$ , leaving only $x = 4$ .
Dividing an equation by an unknown is only safe when you already know that unknown cannot be zero.
How do you solve $(2 x - 3) (3 x + 5) = 0$ ?
If a product is zero then one of its factors must be zero, so set each bracket equal to zero in turn.
That gives $2 x - 3 = 0$ and $3 x + 5 = 0$ , so $x = \frac{3}{2}$ or $x = - \frac{5}{3}$ .
Complete the quadratic formula:
$x = \frac{- b \pm \sqrt{______}}{______}$
The completed formula is:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
The whole of $- b$ sits above the line, and the $2 a$ divides everything above it, including the square root.
Solve $x^{2} - 7 x + 2 = 0$ , giving your answers in exact form.
Substitute $a = 1$ , $b = - 7$ and $c = 2$ into the quadratic formula, putting brackets around the negative:
$x = \frac{7 \pm \sqrt{41}}{2}$
"Exact form" means the answer keeps its surd, so this cannot be simplified any further.
Complete the number of solutions in each case:
$b^{2} - 4 a c > 0$ gives $______$ solution(s).
$b^{2} - 4 a c = 0$ gives $______$ solution(s).
$b^{2} - 4 a c < 0$ gives $______$ solution(s).
The completed rules are:
$b^{2} - 4 a c > 0$ gives two different solutions.
$b^{2} - 4 a c = 0$ gives one repeated solution.
$b^{2} - 4 a c < 0$ gives no solutions.
The count follows from the square root: a positive number has two roots, zero has one, and a negative number has none.
Why is $x^{4} - 3 x^{2} - 4 = 0$ called a hidden quadratic?
Because it has the structure $a {(something)}^{2} + b (something) + c = 0$ , with $x^{2}$ as the something.
Since $x^{4} = {(x^{2})}^{2}$ , the equation is a quadratic in $x^{2}$ rather than in $x$ .
Solve $x^{4} - 3 x^{2} - 4 = 0$ .
Substitute $u = x^{2}$ to get $u^{2} - 3 u - 4 = 0$ , which gives $u = 4$ or $u = - 1$ .
Then $x^{2} = 4$ gives the two solutions $x = \pm 2$ , while $x^{2} = - 1$ gives none, so $x = - 2$ or $x = 2$ .
The solutions of $x^{2} - 3 x - 4 = 0$ are $x = 4$ and $x = - 1$ . Why can you not just add 5 to these to solve ${(x + 5)}^{2} - 3 (x + 5) - 4 = 0$ ?
Substituting $u = x + 5$ gives $u = 4$ or $u = - 1$ , and those are values of $x + 5$ rather than of $x$ .
You therefore subtract 5 from each, giving $x = - 1$ or $x = - 6$ .
Solve $x - \sqrt{x} - 6 = 0$ .
Substitute $u = \sqrt{x}$ to get $u^{2} - u - 6 = 0$ , so $u = 3$ or $u = - 2$ .
Only $u = 3$ is possible, because $\sqrt{x}$ can never be negative, so the single solution is $x = 9$ .
Define identity.
An identity is a statement with an $\equiv$ sign that is true for every value of the variable.
For example $x + x \equiv 2 x$ holds whatever $x$ is, including negatives, fractions and $π$ .
True or False?
$2 x = 10$ is an identity.
False.
The statement is true only when $x = 5$ , so it is an equation rather than an identity.
An identity holds for every value of $x$ , which is why it is written with $\equiv$ instead of $=$ .
What does the method of equating coefficients involve?
Expand and collect like terms on each side, then match the coefficients of each power of $x$ across the identity.
Each power gives its own equation, and those equations can then be solved for the unknown constants.
Complete the identity by equating coefficients:
$3 (x + y) + 2 y \equiv______x +______y$
The completed identity is:
$3 (x + y) + 2 y \equiv 3 x + 5 y$
Expanding the left side gives $3 x + 3 y + 2 y$ , and the two $y$ terms combine to $5 y$ .
Given that $a {(x + 3)}^{2} - b x + 5 \equiv 4 x^{2} + c$ , find $a$ , $b$ and $c$ .
Expanding the left side gives $a x^{2} + 6 a x + 9 a - b x + 5$ , so comparing terms gives $a = 4$ , then $6 a - b = 0$ and $9 a + 5 = c$ .
These solve to give $a = 4$ , $b = 24$ and $c = 41$ .
Why does a missing term still matter when equating coefficients?
Because a term that is absent has a coefficient of zero, and that zero still has to be matched.
If one side has no $x$ term at all, then the coefficients of $x$ on the other side must add up to 0.
Complete the rule for turning a ratio into algebra:
If $x : y$ equals $a : b$ , then $\frac{x}{y} =______$ .
The completed rule is:
If $x : y$ equals $a : b$ , then $\frac{x}{y} = \frac{a}{b}$ .
Writing a ratio as an equation between two fractions is what lets you use algebra on it.
If $A = 6 x$ , $B = x + 1$ and $A : B = 9 : 2$ , find $x$ .
Write the ratio as $\frac{6 x}{x + 1} = \frac{9}{2}$ , then cross-multiply.
That gives $12 x = 9 x + 9$ , so $3 x = 9$ and $x = 3$ .
How do you write " $x %$ of $y$ " and "an increase of $a %$ on $b$ " algebraically?
$x %$ of $y$ is $\frac{x}{100} \times y$ , which simplifies to $\frac{x y}{100}$ .
An increase of $a %$ on $b$ is $b (1 + \frac{a}{100})$ , since the original amount is the 1.
If $A = 6 x$ and $B = x + 1$ , and $A %$ of 50 equals $B %$ of 250, find $x$ .
Write it as $\frac{6 x}{100} \times 50 = \frac{x + 1}{100} \times 250$ , then multiply both sides by 100.
That gives $300 x = 250 x + 250$ , so $50 x = 250$ and $x = 5$ .
Complete the formula for the sum of the interior angles of a polygon with $n$ sides:
$180 ° \times (______)$
The completed formula is:
$180 ° \times (n - 2)$
A hexagon has $n = 6$ , so its interior angles sum to $180 ° \times 4 = 720 °$ .
True or False?
The interior angles of an irregular pentagon still add up to $540 °$ .
True.
The sum depends only on the number of sides, not on whether the polygon is regular.
Only the individual angles change: in a regular pentagon they are all $108 °$ , but in an irregular one they are not.
The interior angles of a pentagon are $3 x + 10$ , $5 x - 8$ , $x$ , $6 x - 2$ and $90$ degrees. Find $x$ .
The angles of a pentagon sum to $540 °$ , so the equation to form is $15 x + 90 = 540$ .
Then $15 x = 450$ , giving $x = 30$ .
Write an expression for the perimeter of a regular pentagon with side length $2 x - 1$ .
The perimeter is $5 (2 x - 1)$ , which expands to $10 x - 5$ .
Regular means all five sides are the same length, so one side length is multiplied by 5.
When forming an expression for a surface area, what is easy to leave out?
The faces that are hidden in the diagram, since only some of a solid's faces are visible from one viewpoint.
Count the faces first, note which of them are identical, then write an area expression for each one.
A cone and a hemisphere have the same radius $r$ and the same total surface area. Using $4 π r^{2}$ for the surface area of a sphere and $π r l$ for the curved surface area of a cone, show that the cone's slant height is $2 r$ .
The hemisphere's total surface area is $\frac{1}{2} \times 4 π r^{2} + π r^{2} = 3 π r^{2}$ , and the cone's is $π r l + π r^{2}$ .
Setting them equal gives $3 π r^{2} = π r (l + r)$ , so $3 r = l + r$ and $l = 2 r$ .

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Counting PrinciplesNext:Binomial Expansion