True or False? A negative discriminant means the quadratic has two negative roots.

False. A negative discriminant means there are no real roots at all, so there is nothing to be negative. The sign of the discriminant counts the roots, and says nothing about whether the roots themselves are positive or negative.

National 5MathsSQAFlashcardsAlgebraic SkillsQuadratic Equations & Roots

Quadratic Equations & Roots (SQA National 5 Maths): Flashcards

Exam code: X847 75

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What must a quadratic equation look like before you try to solve it by factorising?

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Cards in this collection (21)

What must a quadratic equation look like before you try to solve it by factorising?
Rearrange it into the form $a x^{2} + b x + c = 0$ , so that zero is on one side.
It is easier if you also rearrange so that $a$ comes out positive.
Why does setting each bracket equal to zero give the solutions?
If two things multiply together to give zero, then one or the other of them must itself be zero.
So $(x + 4) (x - 1) = 0$ means either $x + 4 = 0$ or $x - 1 = 0$ .
Complete the two solutions by filling in the missing values.
$(x - 3) (x + 7) = 0 \Rightarrow x =______or x =______$
The completed solutions are:
$x = 3 or x = - 7$
When each bracket is just $x$ plus or minus a number, the solutions are those numbers with their signs swapped.
What extra work is needed when a bracket has a number in front of the $x$ ?
Each bracket still gets set equal to zero, but solving it now takes an extra dividing step.
From $(2 x - 3) (3 x + 5) = 0$ the first bracket gives $2 x = 3$ and so $x = \frac{3}{2}$ , and the second gives $x = - \frac{5}{3}$ .
True or False?
$5 x^{2} - x = 0$ can be solved by dividing both sides by $x$ .
False.
Dividing by $x$ throws away the solution $x = 0$ , leaving only one of the two answers.
Factorising to $x (5 x - 1) = 0$ instead keeps both, giving $x = 0$ and $x = \frac{1}{5}$ .
How is a lone $x$ outside a bracket treated when solving?
Treat it as a bracket of its own, so that $x (x - 4) = 0$ reads as $(x) = 0$ or $x - 4 = 0$ .
That gives the two solutions $x = 0$ and $x = 4$ .
How can a pair of solutions be turned back into a factorised form?
Reverse the solving step, so solutions of $- \frac{2}{3}$ and $\frac{1}{2}$ come from the brackets $(3 x + 2)$ and $(2 x - 1)$ .
Check the $x^{2}$ term afterwards, since $12 x^{2} + 2 x - 4$ has the same solutions but needs an extra factor of $2$ outside.
What has to be true of a quadratic equation before the quadratic formula can be used?
The equation must be arranged so that one side reads $= 0$ , giving the form $a x^{2} + b x + c = 0$ .
Anything else has to be rearranged first, or the values of $a$ , $b$ and $c$ will be read off wrongly.
True or False?
Some quadratic equations cannot be solved by factorising but can still be solved by the quadratic formula.
True.
Many quadratics have solutions that are not whole numbers or simple fractions, so no pair of brackets with whole numbers in them will work.
The formula reaches those solutions anyway, which is why $2 x^{2} - 8 x - 3 = 0$ gives $\frac{4 \pm \sqrt{22}}{2}$ .
For $3 x^{2} - 2 x - 4 = 0$ complete the three values that go into the quadratic formula.
$a =______, b =______, c =______$
The completed values are:
$a = 3, b = - 2, c = - 4$
Each coefficient carries the sign written in front of it, so both $b$ and $c$ are negative here.
Why should negative values be put in brackets when substituting into the formula?
Brackets keep the sign attached during the squaring and multiplying, so $b^{2}$ becomes ${(- 2)}^{2} = 4$ rather than $- 4$ .
They matter just as much under the root, where $- 4 \times 3 \times (- 4)$ has to come out as $+ 48$ .
Why does the formula give two solutions rather than one?
The $\pm$ in front of the square root splits the working into two, one using $+$ and one using $-$ .
For $3 x^{2} - 2 x - 4 = 0$ those give $\frac{1 + \sqrt{13}}{3}$ and $\frac{1 - \sqrt{13}}{3}$ .
How do you give the solutions in exact form rather than as decimals?
Work out the number under the root, simplify it as a surd, and then cancel any factor shared with the bottom.
For $2 x^{2} - 8 x - 3 = 0$ the root is $\sqrt{88} = 2 \sqrt{22}$ , giving $\frac{8 \pm 2 \sqrt{22}}{4}$ and then $\frac{4 \pm \sqrt{22}}{2}$ .
What does a calculator answer containing $i$ tell you?
Those solutions are complex rather than real, so they are not what a National 5 question is asking for.
Either the equation was typed in wrongly, or it genuinely has no real solutions.
Define the discriminant of a quadratic.
The discriminant is the expression $b^{2} - 4 a c$ , which is the part of the quadratic formula that sits under the square root.
Its sign tells you how many real roots $a x^{2} + b x + c = 0$ has.
What are the roots of a quadratic function?
The roots are the solutions of the equation you get by setting the function equal to zero.
For $f (x) = a x^{2} + b x + c$ they are the solutions of $a x^{2} + b x + c = 0$ .
Complete the three cases by filling in the missing numbers of roots.
$b^{2} - 4 a c > 0 gives______real and distinct roots$
$b^{2} - 4 a c = 0 gives______repeated real root$
$b^{2} - 4 a c < 0 gives______real roots$
The completed cases are:
$b^{2} - 4 a c > 0 gives two real and distinct roots$
$b^{2} - 4 a c = 0 gives one repeated real root$
$b^{2} - 4 a c < 0 gives no real roots$
One repeated real root may equally be described as two equal real roots.
True or False?
A negative discriminant means the quadratic has two negative roots.
False.
A negative discriminant means there are no real roots at all, so there is nothing to be negative.
The sign of the discriminant counts the roots, and says nothing about whether the roots themselves are positive or negative.
How is the discriminant of $16 x^{2} - 8 x + 1$ worked out?
Read off $a = 16$ , $b = - 8$ and $c = 1$ , then substitute into $b^{2} - 4 a c$ .
That gives ${(- 8)}^{2} - 4 (16) (1) = 64 - 64 = 0$ , so this quadratic has one repeated real root.
Why does a discriminant of $49$ mean two different roots?
A positive discriminant leaves a real square root to be added and subtracted in the quadratic formula, producing two different answers.
A discriminant of $0$ makes that root zero, which is why adding and subtracting it then gives the same answer twice.
What does it tell you when the discriminant is a square number?
The square root comes out exactly, so the quadratic expression can be factorised into brackets with whole numbers in them.
A discriminant of $49$ is a square number, whereas one of $88$ is not, which is why the second needs the formula rather than factorising.

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