True or False? If a quadratic with integer coefficients has a discriminant that is a perfect square , it could have been factorised.

True. A perfect square under the root leaves a whole number, so both solutions are rational and the brackets have integer coefficients. Checking the discriminant first therefore tells you whether factorising would have been the quicker route.

Which function on a graphic display calculator solves a quadratic equation?

A quadratic is solved with the polynomial solver , which on some models sits inside the equation solver menu. You choose a degree or order of 2 for a quadratic, then type in its coefficients.

True or False? If a solution is going to be used in further working, you should round it before carrying on.

False. Keep it in exact form for as long as you can, and round only at the very end. Rounding early makes a small error grow through everything that follows it.

True or False? A calculator that returns solutions containing i has found the answers you want.

False. Those are complex solutions, and they appear precisely when the equation has no real solutions . The answer to give is that the equation cannot be solved, not the values on the display.

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Quadratic Equations (Cambridge (CIE) IGCSE International Maths: Extended): Flashcards

Exam code: 0607

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FrontSolving Quadratics by Factorising
What is a quadratic equation?
FrontSolving Quadratics by Factorising
What does it mean to solve a quadratic equation by factorising?
BackSolving Quadratics by Factorising
Solving a quadratic equation by factorising means:
Making sure it is in the form $a x^{2} + b x + c = 0$ (i.e. with zero on one side).
Factorising the quadratic.
Setting each bracket equal to zero to find the solutions.

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Cards in this collection (21)

What is a quadratic equation?
A quadratic equation is an equation of the form $a x^{2} + b x + c = 0$ ,
where $a$ , $b$ , and $c$ are constants, and $a$ is not equal to zero.
What does it mean to solve a quadratic equation by factorising?
Solving a quadratic equation by factorising means:
1. Making sure it is in the form $a x^{2} + b x + c = 0$ (i.e. with zero on one side).
2. Factorising the quadratic.
3. Setting each bracket equal to zero to find the solutions.
True or False?
If $(x + 4) (x - 1) = 0$ , then either $x + 4 = 0$ or $x - 1 = 0$ .
True.
If $(x + 4) (x - 1) = 0$ , then either $x + 4 = 0$ or $x - 1 = 0$ .
True or false?
The solutions of the equation $(x - 2) (x + 5) = 0$ are $x = - 2$ and $x = 5$ .
False.
To find the solutions of the equation $(x - 2) (x + 5) = 0$ , you should solve the linear equations $x - 2 = 0$ and $x + 5 = 0$ .
The solutions are $x = 2$ and $x = - 5$ .
Note that the signs in front of 2 and 5 are the 'other way round' to how they appear in the brackets.
What two linear equations should you solve to find the solutions of the quadratic equation $(8 x + 7) (2 x - 3) = 0$ ?
To find the solutions of the quadratic equation $(8 x + 7) (2 x - 3) = 0$ , you should solve the linear equations $8 x + 7 = 0$ and $2 x - 3 = 0$ .
The solutions are $x = - \frac{7}{8}$ and $x = \frac{3}{2}$ .
True or false?
$x = 0$ is one of the solutions of the quadratic equation $x (5 x - 1) = 0$ .
True.
If $x (5 x - 1) = 0$ , then either $x = 0$ or $5 x - 1 = 0$ .
So $x = 0$ is one of the solutions.
The other solution is $x = \frac{1}{5}$ .
Before using the quadratic formula, what must the equation look like, and what do you read off it?
The equation must have zero on one side, in the form $a x^{2} + b x + c = 0$ first.
Each coefficient takes the sign in front of it, so in $2 x^{2} - 8 x - 3 = 0$ you have $b = - 8$ and $c = - 3$ here.
True or False?
The brackets can be left out when substituting a negative value of b into the quadratic formula.
False.
Writing $- 8^{2}$ instead of ${(- 8)}^{2}$ squares only the 8 and leaves the result negative.
The discriminant then comes out wrong, so the brackets have to be there.
Define the discriminant of a quadratic equation.
The discriminant is the expression $b^{2} - 4 a c$ found under the square root in the quadratic formula.
Its sign tells you how many solutions the equation has, without having to solve it.
Complete the rule for the number of solutions of a quadratic equation.
If $b^{2} - 4 a c$ is positive there are $______$ different solutions, if it is zero there is $______$ solution, and if it is negative there are $______$ solutions at all.
The completed rule is:
If $b^{2} - 4 a c$ is positive there are two different solutions, if it is zero there is one solution, and if it is negative there are no solutions at all.
The single solution in the middle case is sometimes described as two repeated solutions.
How many solutions does $3 x^{2} - 2 x - 4 = 0$ have?
Its discriminant is ${(- 2)}^{2} - 4 \times 3 \times (- 4) = 4 + 48 = 52$ here.
That value is positive, so the equation has two different solutions.
True or False?
If a quadratic with integer coefficients has a discriminant that is a perfect square, it could have been factorised.
True.
A perfect square under the root leaves a whole number, so both solutions are rational and the brackets have integer coefficients.
Checking the discriminant first therefore tells you whether factorising would have been the quicker route.
Solve $3 x^{2} - 2 x - 4 = 0$ giving exact values in their simplest form.
Substituting into the formula gives $x = \frac{2 \pm \sqrt{52}}{6}$ before any simplifying.
Since $\sqrt{52} = 2 \sqrt{13}$ every term has a factor of 2, which cancels to leave:
$x = \frac{1 \pm \sqrt{13}}{3}$
A quadratic has solutions $4.3452078 \dots$ and $- 0.3452078 \dots$ as decimals. Write each to 3 significant figures.
They become 4.35 and −0.345 to 3 significant figures.
The second keeps an extra decimal place because a leading zero is not a significant figure, so counting starts at the 3.
Which function on a graphic display calculator solves a quadratic equation?
A quadratic is solved with the polynomial solver, which on some models sits inside the equation solver menu.
You choose a degree or order of 2 for a quadratic, then type in its coefficients.
What must you do to $10 x = 19 - 6 x^{2}$ before entering it into the solver?
Rearrange it so that one side is zero, which gives $6 x^{2} + 10 x - 19 = 0$ after collecting terms.
The coefficients to enter are then 6, 10 and −19 in that order.
Complete the sentence about the answers a calculator gives.
A graphic display calculator presents the solutions in $______$ form wherever it can, such as $- 3 + \sqrt{13}$ rather than a long decimal.
The completed sentence is:
A graphic display calculator presents the solutions in exact form wherever it can, such as $- 3 + \sqrt{13}$ rather than a long decimal.
Most models will switch between that form and a decimal at the press of a key.
True or False?
If a solution is going to be used in further working, you should round it before carrying on.
False.
Keep it in exact form for as long as you can, and round only at the very end.
Rounding early makes a small error grow through everything that follows it.
The solver returns $x = \frac{- 5 \pm \sqrt{139}}{6}$ as the solutions. What are they to 4 significant figures?
They are 1.132 and −2.798 to four significant figures.
The two values come from taking $+ \sqrt{139}$ and then $- \sqrt{139}$ in the numerator.
True or False?
A calculator that returns solutions containing i has found the answers you want.
False.
Those are complex solutions, and they appear precisely when the equation has no real solutions.
The answer to give is that the equation cannot be solved, not the values on the display.
A polynomial solver asks for $a_{2}$ and $a_{1}$ and $a_{0}$ as its inputs. What are they?
They are the coefficients of $x^{2}$ and of $x$ and the constant term, matching a, b and c in that order.
For $x^{2} + 6 x - 4 = 0$ they would be 1, 6 and −4 respectively.

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