True or False? The quadratic formula can be used on any quadratic equation, including ones that factorise easily.

True. The formula works on every quadratic equation, so it is never the wrong choice, only sometimes the slower one. If a quadratic factorises, the formula will still deliver those same solutions.

A question asks for solutions correct to 2 decimal places. Which method does that point you towards, and why?

The quadratic formula , because a rounding instruction means the solutions are not whole numbers or simple fractions. That also tells you the quadratic does not factorise neatly, so time spent hunting for brackets would be wasted.

Setting a line and a curve equal gives an equation whose discriminant is zero. What does that tell you about the line?

The line is a tangent to the curve. A zero discriminant means the two graphs meet at exactly one point, which is what being a tangent means.

IGCSEFurther MathsCambridge (CIE)Additional MathsFlashcardsAlgebra & FunctionsQuadratic Functions

Quadratic Functions (Cambridge (CIE) IGCSE Additional Maths): Flashcards

Exam code: 0606

1/52

0Still learning

Know0

FrontSolving Quadratics by Factorising
What form must a quadratic equation be in before you factorise it?

Was this flashcard helpful?

Cards in this collection (52)

What form must a quadratic equation be in before you factorise it?
It must be rearranged to $a x^{2} + b x + c = 0$ , with zero on one side.
The method depends on one side being zero, so factorising an equation such as $x^{2} + 3 x = 10$ before rearranging it will not work.
Why does $(x - 3) (x + 7) = 0$ tell you that $x = 3$ or $x = - 7$ ?
Because if two things multiply to give zero, at least one of them must be zero.
So either $x - 3 = 0$ , giving $x = 3$ , or $x + 7 = 0$ , giving $x = - 7$ .
Notice the solutions come out with the opposite signs to the numbers in the brackets.
Fill in the two solutions:
If $(2 x - 3) (3 x + 5) = 0$ then $x =______$ or $x =______$ .
The completed sentence is:
If $(2 x - 3) (3 x + 5) = 0$ then $x = \frac{3}{2}$ or $x = - \frac{5}{3}$ .
When a bracket has a number in front of $x$ , you still set the bracket to zero, then divide by that number at the end.
True or False?
To solve $x (x - 4) = 0$ , you should start by dividing both sides by $x$ .
False.
Dividing by $x$ throws away the solution $x = 0$ , leaving only $x = 4$ .
Treat the lone $x$ as a bracket of its own, so the two solutions are $x = 0$ and $x = 4$ .
A calculator gives the solutions of $12 x^{2} + 2 x - 4 = 0$ as $x = - \frac{2}{3}$ and $x = \frac{1}{2}$ . Why is $(3 x + 2) (2 x - 1)$ not the full factorisation?
Those brackets expand to $6 x^{2} + x - 2$ , which is only half of the original expression.
The solutions tell you the brackets but not the common factor, so you have to multiply by 2 to get $12 x^{2} + 2 x - 4 = 2 (3 x + 2) (2 x - 1)$ .
How do you solve $x^{2} - 9 = 0$ by factorising?
Recognise it as the difference of two squares and factorise it as $(x + 3) (x - 3) = 0$ .
Setting each bracket to zero gives $x = - 3$ and $x = 3$ .
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 $- b \pm \sqrt{b^{2} - 4 a c}$ sits over $2 a$ , so the $2 a$ divides everything above it, not just the square root.
What must you do to an equation before reading off $a$ , $b$ and $c$ ?
Rearrange it into the form $a x^{2} + b x + c = 0$ , so that one side is zero.
The values of $a$ , $b$ and $c$ are then the coefficients in that arrangement, taken with their signs.
When substituting into the quadratic formula, why put brackets around a negative value?
Because the formula squares $b$ , and the brackets are what keep the sign inside the square.
For $b = - 5$ , writing $(- 5)^{2}$ gives $25$ , but writing $- 5^{2}$ gives $- 25$ and the answer is then wrong.
The formula gives $x = \frac{5 \pm \sqrt{9}}{4}$ . How do you finish?
Work out the $+$ and the $-$ versions separately, because the $\pm$ is what produces the two solutions:
$x = \frac{5 + 3}{4} = 2$
$x = \frac{5 - 3}{4} = \frac{1}{2}$
True or False?
The quadratic formula can be used on any quadratic equation, including ones that factorise easily.
True.
The formula works on every quadratic equation, so it is never the wrong choice, only sometimes the slower one.
If a quadratic factorises, the formula will still deliver those same solutions.
Why must $a$ be non-zero in $a x^{2} + b x + c = 0$ ?
If $a$ were zero there would be no $x^{2}$ term, so the equation would be linear rather than quadratic.
The formula would also break down, because it divides by $2 a$ .
What is the point of writing a quadratic in the form $a (x + p)^{2} + q$ ?
In that form $x$ appears only once, instead of in both an $x^{2}$ term and an $x$ term.
That is what lets you make $x$ the subject and solve the equation directly, by undoing each operation in turn.
For $x^{2} + b x + c$ written as $(x + p)^{2} + q$ , fill in the gaps:
$p$ is $______$ , and $q$ is $______$ .
The completed sentence is:
$p$ is half of $b$ , and $q$ is $c - p^{2}$ .
The $- p^{2}$ is there to cancel the extra $p^{2}$ that appears when $(x + p)^{2}$ is expanded.
How do you start completing the square on $4 x^{2} + 16 x + 5$ ?
Take the factor of 4 out of the $x^{2}$ and $x$ terms only, giving $4 [x^{2} + 4 x] + 5$ .
Complete the square inside the bracket, then expand and simplify:
$4 [(x + 2)^{2} - 4] + 5 = 4 (x + 2)^{2} - 11$
True or False?
When the coefficient of $x^{2}$ is not 1, the constant term must be taken inside the bracket as well.
False.
Only the $x^{2}$ and $x$ terms go inside the bracket; the constant stays outside, where it is untouched.
Taking it inside creates awkward fractions for no benefit.
A question asks you to write an expression in the form $a (x + p)^{2} + q$ but never uses the words "complete the square". What is it asking for?
That is completing the square, written as a target form instead of named.
The form itself is the instruction, so treat it exactly as you would the phrase.
How do you solve $(x + 5)^{2} = 16$ ?
Square root both sides, keeping a $\pm$ sign so that both solutions survive:
$x + 5 = \pm 4$
Then subtract 5 from each version, giving $x = - 1$ and $x = - 9$ .
Why include a $\pm$ sign when you square root both sides?
Because a positive number has two square roots, one positive and one negative, and both give valid solutions.
Writing only the positive root loses one of the two answers.
You are solving $2 x^{2} - 8 x - 24 = 0$ by completing the square. What should you do first?
Divide the whole equation by 2, so that it starts with a plain $x^{2}$ :
$x^{2} - 4 x - 12 = 0$
Dividing is available here because the equation is equal to zero, which is what makes this different from completing the square on an expression.
Which features of a quadratic equation suggest solving it by factorising?
A quadratic that has only two terms, such as $x^{2} - 4 x = 0$ or $x^{2} - 9 = 0$ , since a common factor or the difference of two squares will do the work.
Also any question that asks you to factorise in an earlier part, or whose factorisation you can simply spot.
A question asks for solutions correct to 2 decimal places. Which method does that point you towards, and why?
The quadratic formula, because a rounding instruction means the solutions are not whole numbers or simple fractions.
That also tells you the quadratic does not factorise neatly, so time spent hunting for brackets would be wasted.
The quadratic formula gives $x = \frac{- 4 \pm \sqrt{32}}{4}$ . How do you write this in exact form?
Simplify the surd first, using $\sqrt{32} = 4 \sqrt{2}$ :
$x = \frac{- 4 \pm 4 \sqrt{2}}{4} = - 1 \pm \sqrt{2}$
Every term shares a factor of 4, which is what makes the fraction cancel completely.
Define a hidden quadratic equation.
A hidden quadratic is an equation with the structure $a (something)^{2} + b (something) + c = 0$ , where the something is not simply $x$ .
For example $x^{4} - 3 x^{2} - 4 = 0$ is a quadratic in $x^{2}$ , because $x^{4} = (x^{2})^{2}$ .
How do you solve a hidden quadratic equation?
Substitute a single letter for the repeated expression, so that $u = something$ turns the equation into $a u^{2} + b u + c = 0$ .
Solve that easier quadratic, then replace $u$ with what it stood for and solve each resulting equation separately.
Those final equations may have two solutions, one, or none at all.
Fill in the substitution that turns each equation into a quadratic:
$x - 3 \sqrt{x} - 4 = 0$ is a quadratic in $______$ , and $4^{x} - 3 \times 2^{x} - 4 = 0$ is a quadratic in $______$ .
The completed sentence is:
$x - 3 \sqrt{x} - 4 = 0$ is a quadratic in $\sqrt{x}$ , and $4^{x} - 3 \times 2^{x} - 4 = 0$ is a quadratic in $2^{x}$ .
The second one needs a change of base first, since $4^{x} = (2^{2})^{x} = (2^{x})^{2}$ .
Solving $x - \sqrt{x} - 6 = 0$ by substitution gives $\sqrt{x} = 3$ or $\sqrt{x} = - 2$ . What are the solutions in $x$ ?
Only $x = 9$ , from $\sqrt{x} = 3$ .
The branch $\sqrt{x} = - 2$ gives nothing, because a square root is never negative.
This is why each substituted equation has to be solved and checked in its own right.
True or False?
If $x^{2} - 3 x - 4 = 0$ has solutions $x = 4$ and $x = - 1$ , then $(x + 5)^{2} - 3 (x + 5) - 4 = 0$ has solutions $x = 9$ and $x = 4$ .
False.
Substituting $u = x + 5$ gives $u = 4$ or $u = - 1$ , and then $x = u - 5$ , so the solutions are $x = - 1$ and $x = - 6$ .
You subtract 5 rather than adding it, which is why skipping the substitution is dangerous.
Define the discriminant of a quadratic.
The discriminant of $a x^{2} + b x + c$ is $b^{2} - 4 a c$ , the part of the quadratic formula that sits under the square root.
It is often written as the Greek capital delta, $Δ$ .
How does the value of the discriminant decide the number of real roots?
If $b^{2} - 4 a c > 0$ there are two different real roots, because the square root gives a non-zero amount to add and subtract.
If $b^{2} - 4 a c = 0$ there is one repeated root, because adding and subtracting zero gives the same answer twice.
If $b^{2} - 4 a c < 0$ there are no real roots, because you cannot square root a negative number.
True or False?
A quadratic equation can have no real roots at all.
True.
When $b^{2} - 4 a c < 0$ there are no real solutions, and its graph simply never reaches the $x$ -axis.
A negative discriminant is a valid answer, not a sign that you have made a mistake.
Fill in the gaps linking the discriminant to the graph:
If $b^{2} - 4 a c > 0$ the graph crosses the $x$ -axis $______$ ; if $b^{2} - 4 a c = 0$ it $______$ the $x$ -axis; if $b^{2} - 4 a c < 0$ it $______$ the $x$ -axis.
The completed sentence is:
If $b^{2} - 4 a c > 0$ the graph crosses the $x$ -axis twice; if $b^{2} - 4 a c = 0$ it touches the $x$ -axis; if $b^{2} - 4 a c < 0$ it never reaches the $x$ -axis.
Touching rather than crossing is what a repeated root looks like on a graph.
How do you find the number of intersections between a curve $y = f (x)$ and a line $y = g (x)$ ?
Set them equal, rearrange to the form $a x^{2} + b x + c = 0$ , and find the discriminant of that equation.
Its sign gives the answer: positive means the graphs meet twice, zero means once, and negative means they do not meet.
Setting a line and a curve equal gives an equation whose discriminant is zero. What does that tell you about the line?
The line is a tangent to the curve.
A zero discriminant means the two graphs meet at exactly one point, which is what being a tangent means.
To show $y = 3 x - 7$ is a tangent to $y = (x + 3) (x - 2)$ , you reach $x^{2} - 2 x + 1 = 0$ . How do you finish?
Find the discriminant and show it is zero:
$Δ = (- 2)^{2} - 4 (1) (1) = 0$
Since the discriminant is zero the graphs meet exactly once, so the line is a tangent.
Before finding a discriminant in terms of an unknown constant, what must you do to the equation?
Rearrange it so that one side is zero, then read off $a$ , $b$ and $c$ with their signs before substituting.
Identifying the three coefficients as a separate step is what keeps the signs straight when they are expressions rather than numbers.
Define a parabola.
A parabola is the curve made by a quadratic graph $y = a x^{2} + b x + c$ .
It is symmetrical, and it is shaped either like a $\cup$ or like a $\cap$ .
What decides whether a quadratic graph is $\cup$ -shaped or $\cap$ -shaped?
The sign of $a$ , the coefficient of $x^{2}$ .
A positive $a$ gives a $\cup$ -shape with a minimum point, and a negative $a$ gives a $\cap$ -shape with a maximum point.
True or False?
Every quadratic graph crosses the $y$ -axis.
True.
Substituting $x = 0$ into $a x^{2} + b x + c$ always gives the value $c$ , so the graph passes through $(0, c)$ .
The $x$ -axis is the one it may miss, since a quadratic can have two roots, one, or none.
How do you find the roots of a quadratic graph, and what are they on the sketch?
Set $y = 0$ and solve $a x^{2} + b x + c = 0$ .
The roots are the values of $x$ where the curve meets the $x$ -axis, so on a sketch they are the $x$ -intercepts.
Fill in the coordinates of the turning point:
If a quadratic is written as $a (x + p)^{2} + q$ , its turning point is at $(______,______)$ .
The completed sentence is:
If a quadratic is written as $a (x + p)^{2} + q$ , its turning point is at $(- p, q)$ .
Watch the sign: the $x$ -coordinate is the negative of the number inside the bracket.
Why does $a (x + p)^{2} + q$ give the turning point straight away?
Because a squared bracket can never be negative, so $(x + p)^{2}$ is smallest when it equals zero.
That happens at $x = - p$ , and there the expression is just $q$ , so $(- p, q)$ is the turning point.
$f (x) = 2 x^{2} - 4 x - 6$ has turning point $(1, - 8)$ . What is the range of $f (x)$ ?
The range is $f (x) \geq - 8$ .
The graph is $\cup$ -shaped, so $- 8$ is the smallest value it reaches, and it does reach it, at $x = 1$ .
True or False?
A quadratic with a maximum point at $(2, 5)$ has range $f (x) < 5$ .
False.
The range is $f (x) \leq 5$ , because the value 5 is actually reached, at $x = 2$ .
A turning point belongs to the graph, so its $y$ -value is always included in the range.
Why should you rearrange a quadratic inequality so that the $x^{2}$ term is positive?
Because you solve it by picturing the graph, and a positive $x^{2}$ term guarantees a $\cup$ -shape.
If you leave it negative the curve is the other way up and the region you want flips, which is where most sign errors come from.
What must you do to an inequality if you multiply or divide it by a negative number?
Flip the inequality sign, so $<$ becomes $>$ and $\geq$ becomes $\leq$ .
It is safer to rearrange so that this never comes up, by moving terms to whichever side keeps the $x^{2}$ term positive.
A $\cup$ -shaped quadratic has roots $x_{1}$ and $x_{2}$ , where $x_{1} \leq x_{2}$ . Fill in the gaps:
The solution of $a x^{2} + b x + c < 0$ is $______$ , and the solution of $a x^{2} + b x + c > 0$ is $______$ .
The completed sentence is:
The solution of $a x^{2} + b x + c < 0$ is $x_{1} < x < x_{2}$ , and the solution of $a x^{2} + b x + c > 0$ is $x < x_{1}$ or $x > x_{2}$ .
Below the axis is the single stretch between the roots; above the axis is the two arms outside them.
True or False?
The solution $k < - 2$ or $k > 3$ can equally well be written as $k < - 2$ and $k > 3$ .
False.
Or means a value only has to satisfy one of the two, which is right for the two separate arms of a $\cup$ -shaped curve.
And would demand both at once, and no number is both less than $- 2$ and greater than 3, so it describes nothing.
A question asks for the values of $k$ for which an equation has no real roots. How does that become a quadratic inequality?
No real roots means the discriminant is negative, so you write $b^{2} - 4 a c < 0$ with $k$ inside it.
Simplifying that gives a quadratic inequality in $k$ , which you then solve by sketching in the usual way.
How do you recognise the kind of inequality whose answer is a region on a graph?
By the number of variables: these inequalities involve both $x$ and $y$ , and several are usually given at once.
The answer is then an area of the graph rather than a set of values of $x$ , and it is normally labelled $R$ .
Fill in the gaps about boundary lines on an inequality graph:
A strict inequality such as $<$ or $>$ is drawn with a $______$ line, while $\leq$ or $\geq$ is drawn with a $______$ line.
The completed sentence is:
A strict inequality such as $<$ or $>$ is drawn with a dotted line, while $\leq$ or $\geq$ is drawn with a solid line.
The solid line shows that the points on the boundary itself satisfy the inequality, and the dotted line shows that they do not.
You have drawn a boundary curve and need to know which side satisfies the inequality. What is the quickest way to decide?
Pick a point clearly on one side and substitute it into the inequality.
The origin is usually easiest: if $(0, 0)$ satisfies $y < x^{2} + 1$ , then the whole side containing the origin is the side you want.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:FunctionsNext:Factors of Polynomials