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Either paper 2 marks
Here is the graph of \(y = 0.5x^2 + 3x - 2\)
Use the graph to estimate the solutions of \(0.5x^2 + 3x - 2 = 0\)
Show answer
\(x \approx -6.6\) and \(x \approx 0.6\)The solutions of \(0.5x^2 + 3x - 2 = 0\) are the values of \(x\) where \(y = 0\), so find where the curve meets the line \(y = 0\).
The curve meets it between \(x = -7\) and \(x = -6\), at about \(x \approx -6.6\).
It meets it again between \(x = 0\) and \(x = 1\), at about \(x \approx 0.6\).
The solutions are \(x \approx -6.6\) and \(x \approx 0.6\).
- B1smaller solution \(x \approx -6.6\) (within 0.2)
- B1larger solution \(x \approx 0.6\) (within 0.2)
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Either paper 2 marks
Solve \(x^2 + 6x + 5 = 0\)
Show answer
\(x = -5\) or \(x = -1\)Look for two numbers that multiply to give 5 and add to give 6. They are 5 and 1, so the left-hand side factorises:
\((x + 5)(x + 1) = 0\)
A product is zero only when one of the brackets is zero.
\(x + 5 = 0\) gives \(x = -5\), and \(x + 1 = 0\) gives \(x = -1\).
- M1factorises: \((x + 5)(x + 1) = 0\)
- A1solutions \(x = -5\) and \(x = -1\)
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Either paper 2 marks
\(x = 3\) is a solution of the equation
\(x^2 +x + c = 0\)
Work out the value of \(c\).
Show answer
\(-12\)A solution is a value that makes the equation true, so substitute \(x = 3\) and the left-hand side must come to zero.
\(3^2 + \times 3 + c = 0\)
\(9 + 3 + c = 0\)
\(9 + 3 = 9 + 3\), so \(c = -12\)
- M1substitutes \(x = 3\) to get \(9 + 3 + c = 0\)
- A1\(c = -12\)
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Either paper 4 marks
Meera is using a graphical method to solve \(x^2 - 4x = 0\)
She draws the graph of \(y = x^2\) and a straight line on the same grid.
Here is the graph of \(y = x^2\)
(a)Complete her method to solve \(x^2 - 4x = 0\)
Write down the equation of the straight line she draws and use the graph to find the solutions.
(b)Levi is solving \(5x^2 - 8x = 0\)
He uses this method.
\(5x^2 - 8x = 0\)
Add \(8x\) to both sides:
\(5x^2 = 8x\)
Divide both sides by \(x\):
\(5x = 8\)
Divide both sides by \(5\):
\(x = 1.6\)
Evaluate his method and his answer.
Show answer
(a) the line is \(y = 4x\); the solutions are \(x = 0\) and \(x = 4\)
(b) Levi is wrong to divide by \(x\), because \(x\) can be \(0\), so he loses the solution \(x = 0\); his \(x = 1.6\) is a solution but the full answer is \(x = 0\) and \(x = 1.6\)Rearrange so that one side is the drawn curve:
\(x^2 = 4x\)
The left-hand side is the curve \(y = x^2\) that is already drawn, so the straight line to draw is \(y = 4x\).
The solutions are the \(x\)-values where the line and the curve meet. The line passes through the origin, and at \(x = 4\) it is at height \(16\), which is on the curve. So the graphs meet where
\(x = 0\) and \(x = 4\)
Check \(x = 0\) in the equation: \(5 \times 0^2 - 8 \times 0 = 0\), so \(x = 0\) is a solution.
Dividing both sides by \(x\) is only allowed when \(x\) is not zero. Here \(x\) could be zero, so the division throws that solution away. This is the same lesson as in part (a), where the curve and the line meet at the origin as well as at a second point.
His value \(x = 1.6\) does satisfy the equation, so his answer is correct as far as it goes, but it is incomplete.
A safe method is to factorise: \(x(5x - 8) = 0\), which gives
\(x = 0\) and \(x = 1.6\)
- (a) B1the line \(y = 4x\)
- (a) A1both solutions, \(x = 0\) and \(x = 4\)
- (b) M1dividing by \(x\) loses the solution \(x = 0\), which makes both sides zero
- (b) A1both solutions, \(x = 0\) and \(x = 1.6\), so his answer is incomplete
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Either paper 3 marks
Solve \(x^2 = 9x - 14\)
Show answer
\(x = 2\) or \(x = 7\)A quadratic can only be factorised once one side of it is zero, so move every term to the left-hand side:
\(x^2 - 9x + 14 = 0\)
Now look for two numbers that multiply to give 14 and add to give -9. They are -2 and -7:
\((x - 2)(x - 7) = 0\)
So \(x = 2\) or \(x = 7\).
- M1rearranges to \(x^2 - 9x + 14 = 0\)
- M1factorises \((x - 2)(x - 7) = 0\)
- A1\(x = 2\) or \(x = 7\)
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Either paper 4 marks
The width of a rectangle is \(x\) cm.
The length of the rectangle is 2 cm more than its width.
The area of the rectangle is 63 cm².
Find the width of the rectangle.
Show answer
\(7\ \text{cm}\)Let \(x\) stand for the width in centimetres. The length is then \(x + 2\) cm, and area is length \(\times\) width:
\(x(x + 2) = 63\)
\(x^2 + 2x = 63\)
\(x^2 + 2x - 63 = 0\)
Two numbers that multiply to give -63 and add to give 2: they are -7 and 9.
\((x - 7)(x + 9) = 0\)
\(x = 7\) or \(x = -9\)
A width cannot be negative, so \(x = -9\) is rejected.
The width is 7 cm.
- P1forms \(x(x + 2) = 63\)
- M1rearranges to \(x^2 + 2x - 63 = 0\)
- M1factorises \((x - 7)(x + 9) = 0\)
- A1width \(7\) cm, rejecting \(-9\)
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Either paper 3 marks
The solutions of the equation
\(x^2 + bx + c = 0\)
are \(x = 2\) and \(x = 8\).
Calculate the value of \(b\) and the value of \(c\).
Show answer
\(b = -10\) and \(c = 16\)If the solutions are \(2\) and \(8\), the equation factorises as
\((x - 2)(x - 8) = 0\)
Multiplying that out gives \(x^2 - 10x + 16\), so
\(b = -10\) and \(c = 16\)
- M1factorised form \((x - 2)(x - 8) = 0\)
- A1\(b = -10\)
- A1\(c = 16\)
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Either paper 5 marks
The diagram shows the plan of the floor of an L-shaped room. All lengths are in metres.
The area of the floor is 169 m².
(a)Show that \(x^2 + 8x - 153 = 0\)
(b)By factorising \(x^2 + 8x - 153\), work out the value of \(x\). You must show your working.
Show answer
(a) \(x(x + 4) + 4(x + 4) = 169\), so \(x^2 + 8x - 153 = 0\)
(b) \(9\ \text{m}\)Split the floor with a horizontal line into a lower rectangle and an upper rectangle.
Lower rectangle: \((x + 4) \times x = x^2 + 4x\)
Upper rectangle: \(4 \times (x + 4) = 4x + 16\)
The two areas add up to the area of the floor:
\(x^2 + 4x + 4x + 16 = 169\)
Collect the terms in \(x\):
\(x^2 + 8x + 16 = 169\)
Subtract 169 from both sides:
\(x^2 + 8x - 153 = 0\)
Two numbers that multiply to give -153 and add to give 8: they are -9 and 17.
\((x - 9)(x + 17) = 0\)
\(x = 9\) or \(x = -17\)
A length cannot be negative: if \(x = -17\) then the side of length \(x\) would be -17 m, so \(x = -17\) is rejected.
\(x = 9\) m
- (a) M1area of the lower rectangle \(x(x + 4) = x^2 + 4x\)
- (a) M1area of the upper rectangle \(4(x + 4) = 4x + 16\)
- (a) A1sum set equal to \(169\) and rearranged to \(x^2 + 8x - 153 = 0\)
- (b) M1factorises to \((x - 9)(x + 17) = 0\)
- (b) A1\(x = 9\), with \(x = -17\) rejected as a negative length
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Either paper 4 marks
Jo wants to work out the solutions of \(x^2 + 9x + 2 = 0\)
She says, "This equation has no solutions."
(a)Is Jo correct? Tick a box.
\(\square\) Yes \(\square\) No
Give a reason for your answer.
(b)Without expanding any brackets, show how to work out the exact solutions of
\(9(x + 2)^2 = 25\)
Give the solutions.
Show answer
(a) No. Here \(b^2 - 4ac = 73\), which is positive, so the equation has two solutions, found by the quadratic formula or by completing the square.
(b) \(x = -\frac{1}{3}\) or \(x = -\frac{11}{3}\)Whether there are solutions depends on the sign of the discriminant, so work it out.
With \(a = 1\), \(b = 9\) and \(c = 2\):
\(b^2 - 4ac = (9)^2 - 4 \times 1 \times (2) = 81 - 8 = 73\)
The discriminant is positive, so the equation has two solutions, found by the quadratic formula or by completing the square.
So Jo is not correct.
Divide both sides by \(9\):
\((x + 2)^2 = \frac{25}{9}\)
Take the square root of both sides, remembering both signs:
\(x + 2 = \pm \frac{5}{3}\)
Subtract \(2\) from both sides:
\(x = -2 \pm \frac{5}{3}\)
\(x = -\frac{1}{3}\) or \(x = -\frac{11}{3}\)
- (a) A1No, Jo is not correct
- (a) C1reason: \(b^2 - 4ac = 73\) is positive, so the equation has two solutions, found by the quadratic formula or by completing the square
- (b) M1divides by \(9\) and square roots with both signs: \(x + 2 = \pm \frac{5}{3}\)
- (b) A1both solutions \(-\frac{1}{3}\) and \(-\frac{11}{3}\)
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Either paper 2 marks
Isla is asked to solve \(4t^2 + 5t - 6 = 0\)
Here is her working and answer.
\((4t - 3)(t + 2) = 0\)
\(t = \frac{3}{4}\) or \(t = 2\)
(a)What mistake has Isla made?
(b)Work out the correct solutions of the equation.
Show answer
(a) Isla got the sign of the root wrong in the bracket with just t. The bracket that Isla solved to get \(t = 2\) gives \(t = -2\).
(b) \(t = -2\) or \(t = \frac{3}{4}\)Solve each bracket on its own as a linear equation.
\(4t - 3 = 0\) gives \(4t = 3\), so \(t = \frac{3}{4}\).
\(t + 2 = 0\) gives \(t = -2\).
Check Isla's answers by substituting each into its bracket. Putting \(t = 2\) into the bracket it came from gives \(4\), not 0, so that root is wrong.
Isla got the sign of the root wrong in the bracket with just t.
Set each bracket equal to zero and solve it.
\(4t - 3 = 0\) gives \(4t = 3\), so \(t = \frac{3}{4}\).
\(t + 2 = 0\) gives \(t = -2\).
The solutions are \(t = -2\) and \(t = \frac{3}{4}\).