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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 \(0.5x^2 + 3x - 2 = 0\)
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\(x \approx -6.6\) 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 \(x^2 + 6x + 5 = 0\)
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\(x = -5\) 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 \(x = -1\).
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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 \(c\).
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\(-12\)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 \(c = -12\)
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Either paper 3 marks
The solutions of the equation
\(x^2 + bx + c = 0\)
are \(x = -6\) and \(x = -9\).
Work out the value of \(b\) and the value of \(c\).
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\(b = 15\) and \(c = 54\)If the solutions are \(-6\) and \(-9\), the equation factorises as
\((x + 6)(x + 9) = 0\)
Multiplying that out gives \(x^2 + 15x + 54\), so
\(b = 15\) and \(c = 54\)
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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 244 m².
(a)Show that \(x^2 + 8x - 240 = 0\)
(b)By factorising \(x^2 + 8x - 240\), work out the value of \(x\). You must show your working.
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(a) \(x(x + 4) + 4(x + 1) = 244\), so \(x^2 + 8x - 240 = 0\)
(b) \(12\ \text{m}\)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: \(4 \times (x + 1) = 4x + 4\)
The two areas add up to the area of the floor:
\(x^2 + 4x + 4x + 4 = 244\)
Collect the terms in \(x\):
\(x^2 + 8x + 4 = 244\)
Subtract 244 from both sides:
\(x^2 + 8x - 240 = 0\)
Two numbers that multiply to give -240 and add to give 8: they are -12 and 20.
\((x - 12)(x + 20) = 0\)
\(x = 12\) or \(x = -20\)
A length cannot be negative: if \(x = -20\) then the side of length \(x\) would be -20 m, so \(x = -20\) is rejected.
\(x = 12\) m
- (a) M1area of the lower rectangle \(x(x + 4) = x^2 + 4x\)
- (a) M1area of the upper rectangle \(4(x + 1) = 4x + 4\)
- (a) A1sum set equal to \(244\) and rearranged to \(x^2 + 8x - 240 = 0\)
- (b) M1factorises to \((x - 12)(x + 20) = 0\)
- (b) A1\(x = 12\), with \(x = -20\) rejected as a negative length
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Either paper 2 marks
Leila is asked to solve \(4t^2 - 17t + 15 = 0\)
Here is her working and answer.
\((t - 3)(4t - 5) = 0\)
\(t = -3\) or \(t = \frac{5}{4}\)
(a)What mistake has Leila made?
(b)Work out the correct solutions of the equation.
Show answer
(a) Leila got the sign of the root wrong in the bracket with just t. The bracket that Leila solved to get \(t = -3\) gives \(t = 3\).
(b) \(t = \frac{5}{4}\) or \(t = 3\)Solve each bracket on its own as a linear equation.
\(4t - 5 = 0\) gives \(4t = 5\), so \(t = \frac{5}{4}\).
\(t - 3 = 0\) gives \(t = 3\).
Check Leila's answers by substituting each into its bracket. Putting \(t = -3\) into the bracket it came from gives \(-6\), not 0, so that root is wrong.
Leila got the sign of the root wrong in the bracket with just t.
Set each bracket equal to zero and solve it.
\(4t - 5 = 0\) gives \(4t = 5\), so \(t = \frac{5}{4}\).
\(t - 3 = 0\) gives \(t = 3\).
The solutions are \(t = \frac{5}{4}\) and \(t = 3\).
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Either paper 2 marks
Here is the graph of \(y = x^2 + 7x + 9\)
Use the graph to estimate the solutions of \(x^2 + 7x + 9 = 0\)
Show answer
\(x \approx -5.3\) and \(x \approx -1.7\)The solutions of \(x^2 + 7x + 9 = 0\) are the values of \(x\) where \(y = 0\), so find where the curve meets the line \(y = 0\).
The curve meets it between \(x = -6\) and \(x = -5\), at about \(x \approx -5.3\).
It meets it again between \(x = -2\) and \(x = -1\), at about \(x \approx -1.7\).
The solutions are \(x \approx -5.3\) and \(x \approx -1.7\).
- B1smaller solution \(x \approx -5.3\) (within 0.2)
- B1larger solution \(x \approx -1.7\) (within 0.2)
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Either paper 2 marks
\(x^2 - 11x + 18 = 0\)
Work out the values of \(x\).
Show answer
\(x = 2\) or \(x = 9\)Look for two numbers that multiply to give 18 and add to give -11. They are -2 and -9, so the left-hand side factorises:
\((x - 2)(x - 9) = 0\)
A product is zero only when one of the brackets is zero.
\(x - 2 = 0\) gives \(x = 2\), and \(x - 9 = 0\) gives \(x = 9\).
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Either paper 2 marks
\(x = 1\) is a solution of the equation
\(x^2 - 6x + c = 0\)
Calculate the value of \(c\).
Show answer
\(5\)A solution is a value that makes the equation true, so substitute \(x = 1\) and the left-hand side must come to zero.
\(1^2 - 6 \times 1 + c = 0\)
\(1 - 6 + c = 0\)
\(1 - 6 = 1 - 6\), so \(c = 5\)
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Either paper 3 marks
The solutions of the equation
\(x^2 + bx + c = 0\)
are \(x = -2\) and \(x = 9\).
Work out the value of \(b\) and the value of \(c\).
Show answer
\(b = -7\) and \(c = -18\)If the solutions are \(-2\) and \(9\), the equation factorises as
\((x + 2)(x - 9) = 0\)
Multiplying that out gives \(x^2 - 7x - 18\), so
\(b = -7\) and \(c = -18\)