Formation of quadratics
Sometimes we have to form quadratic equations to help us solve problems.
Example
Dave has a box which has one side of length (\({x}\) – 3) and another side of length (\({x}\) + 2). Find an expression for the area of Dave’s box.
Solution
To solve this problem we must know that the area of a box is the width multiplied by the height. If we know that, we can say:
Area = (\({x}\) – 3)(\({x}\) + 2).
Expanding the brackets gives us \({x^2}\) -3\({x}\) + 2\({x}\) - 6 = \({x^2}\) – \({x}\) – 6
We have answered the question by forming the quadratic \({x^2}\) – \({x}\) – 6 which is an expression for the area of the box.
Question
A square with sides each equal to (\({x}\) + 3) cm has a smaller square of area 5 cm2 cut out of it. Write an expression for the area of the remaining shape.
Area of square = (\({x}\) + 3)(\({x}\) + 3) = \({x^2}\) + 3\({x}\) + 3\({x}\) + 9 = x2 + 6\({x}\) + 9
After removing the smaller square:
Area remaining will be = \({x^2}\) + 6\({x}\) + 9 – 5 = \({x^2}\) + 6\({x}\) + 4
More guides on this topic
- Equations of lines – WJEC
- Equations of curves - Intermediate & Higher tier – WJEC
- Basic algebra – WJEC
- Equations and formulae – WJEC
- Factorising - Intermediate & Higher tier – WJEC
- Sequences – WJEC
- Functions - Higher only – WJEC
- Inequalities - Intermediate & Higher tier – WJEC
- Simultaneous equations - Intermediate & Higher tier – WJEC