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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 (6)

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 $x = \frac{1}{5}$ .

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