**Quadratics**

Calculators are allowed for all questions unless indicated by the question. There are 5 questions for this topic. The difficulty increases with the question number.

**Question 1:** (Non-calculator)

$x^2 -x -2 = 0$

A. Solve this equation for $x$. (3 marks)

B. Calculate both values of $y$, where $y = 1.2x$ (2 marks)

Answer to part A

Using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ (1 mark for using the correct formula)

$a = 1, b= -1, c = -2$

Subbing this into the quadratic formula: ($b^2 – 4ac = (-1)^2 – 4(1)(-2) = 9$) (1 mark)

$x=\frac{1\pm\sqrt{9}}{2}$

$x = -1, x = 2$ (1 mark for both solutions)

Answer to part B

For the solution where $x = -1$, $y = 1.2(-1) = -1.2$ (1 mark)

When $x = 2$, $y = 1.2(2) = 2.4$ (1 mark)

**Question 2:** (Non-calculator)

Solve the following quadratic equation:

$x^2 -1 = 0 $ (3 marks)

Answer to part A

Method 1, Using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

We know that there is an $x^2$ which means $a=1$,

We have no $x$ terms (that aren’t squared) so $b = 0$

We have a constant of -1, so $c=-1$ (1 mark for recognising that $a=1, b=0, c=-1$)

Subbing these values into the quadratic formula: $x=\frac{\pm\sqrt{-4\times 1\times (-1)}}{2\times 1}$

Giving $x = 1$, or $x = -1$, we have 2 solutions because of the $\pm$ sign inside the fraction. (2 marks for both solutions, 1 mark for 1 correct solution)

Method 2, Observation:

The equation can be re-arranged to $x^2 = 1$, the only numbers that when squared $=1$ are $1$ and $-1$ ($(-1)\times(-1) = 1$) (1 mark)

So the solutions are $1$ and $-1$ (1 mark)

**Question 3:**

$x^2 + x -2 = 0$

A. Factorise this quadratic. (2 marks)

B. Without using the quadratic formula, calculate the values of $x$ (2 marks)

Answer to part A

Factorise this quadratic by finding 2 numbers that add to make 1, and multiply to make -2.

These numbers are -1, and 2. (1 mark)

$= (x-1)(x+2)$ (1 mark)

Answer to part B

Now that the equation can be factorised to $(x-1)(x+2)$, the solutions are when each bracket is equal to 0.

$x-1=0$, x = 1 $ (1 mark)

$x+2 = 0, x = -2$ (1 mark)

**Question 4:**

Given that the area of the rectangle is 40

A. Calculate the value of $x$

Answer to part A

The area of the rectangle is $(2x)(2x-6) = 40$

Expanding the brackets gives: $4x^2 -12x = 40$

$4x^2 -12x – 40 = 0$ (1 mark)

Using the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$a = 4, b = -12, c= -40$

$x=\frac{12\pm\sqrt{144-(4\times 4 \times (-40))}}{8}$

$x = -2, x = 5$ (1 mark)

Ignore negative solution as the length of each side cannot be negative.

$x = 5$ (1 mark)

**Question 5:**

Given the following equation:

$y = 2x^2 -4x + 2$

A. Solve for $x$ when $y=0$ (2 marks)

B. Sketch this graph, labelling the point of intersection with the $y$ and $x$ axis (3 marks)

Answer to part A

Using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ (1 mark for using the correct formula)

$a = 2, b = -4, c = 2$

$x=\frac{4\pm\sqrt{16-(4\times 2 \times 2)}}{4}$

$x = 1$ (repeated root) (1 mark)

Answer to part B

To calculate where the graph touches the $y$ axis, you must calculate the value of $y$ when $x=0 $

$y = 2$ when $x=0$

Award 1 mark for correct shape, 1 mark for correct $y$ intersect co-ordinates, 1 mark for correct $x$ axis intersect co-ordinates.