Solving Quadratics

Topic

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Higher June 2025 Paper 1 Q22

EdexcelCurrent spec5 marksSimultaneous Equations Solving Quadratics

22 Solve the simultaneous equations

\[\begin{aligned} x^2 + 3y + y^2 &= 7 \\ y &= x + 2 \end{aligned}\]

Show clear algebraic working.

(5)

Mark scheme

Scheme	Marks
eg \(x^2 + 3(x + 2) + (x + 2)^2 = 7\) or eg \((y - 2)^2 + 3y + y^2 = 7\)	M1
eg \(2x^2 + 7x + 3 (= 0)\) \(2x^2 + 7x = -3\) or eg \(2y^2 - y - 3 (= 0)\) \(2y^2 - y = 3\)	M1ft
eg \((2x + 1)(x + 3)(= 0)\) or \(x = \dfrac{-7 \pm \sqrt{(7)^2 - 4 \times 2 \times 3}}{2 \times 2}\) or \(\left(x + \dfrac{7}{4}\right)^2 - \left(\dfrac{7}{4}\right)^2 = -\dfrac{3}{2}\) \(\left(x = -\dfrac{1}{2} \text{ and } x = -3\right)\) or eg \((y + 1)(2y - 3)(= 0)\) or \(y = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times -3}}{2 \times 2}\) or \(\left(y - \dfrac{1}{4}\right)^2 - \left(\dfrac{1}{4}\right)^2 = \dfrac{3}{2}\) \(\left(y = -1 \text{ and } y = \dfrac{3}{2}\right)\)	M1ft
\(y = \text{``}{-\dfrac{1}{2}}\text{''} + 2\left(= \dfrac{3}{2}\right)\) and \(y = \text{``}{-3}\text{''} + 2\;(= -1)\) or \(x = \text{``}{-1}\text{''} - 2\;(= -3)\) and \(x = \text{``}{\dfrac{3}{2}}\text{''} - 2\left(= -\dfrac{1}{2}\right)\)	M1ft
Working required Answer: \(x = -\dfrac{1}{2}\) and \(y = \dfrac{3}{2}\) \(x = -3\) and \(y = -1\)	A1
	(5)
	(5 marks)

Notes

M1: for substitution of \(y = x + 2\) (or \(x = \pm y \pm 2\)) into \(x^2 + 3y + y^2 = 7\) to obtain an equation in \(x\) only (or \(y\) only)

M1ft: dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of \(ax^2 + bx + c\;(= 0)\) where at least 2 coefficients (\(a\) or \(b\) or \(c\)) are correct

M1ft: dep on first M1 method to solve their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as eg \(\dfrac{-7 \pm \sqrt{49 - 24}}{4}\) or \(\dfrac{1 \pm \sqrt{1 + 24}}{4}\)

or if factorising allow brackets which expanded give 2 out of 3 terms correct)

or correct values for \(x\)

or correct values for \(y\)

M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations
or fully correct values for the other variable (correct labels for \(x\) / \(y\))
or for one correct pair of values

A1: oe dep on M2 (allow coordinates)

If they find the values of \(y\) but think they are the values of \(x\) then the maximum mark is 3

Higher June 2025 Paper 1R Q22

EdexcelCurrent spec5 marksSimultaneous Equations Solving Quadratics

22 Solve the simultaneous equations

\[\begin{aligned} x^2 + y^2 &= 41 \\ 2x + y &= 3 \end{aligned}\]

Show clear algebraic working.

(5)

Mark scheme

Scheme	Marks
\(x^2 + (-2x + 3)^2 = 41\) or \(\left(\dfrac{-y + 3}{2}\right)^2 + y^2 = 41\)	M1
eg \(5x^2 - 12x - 32 (= 0)\) oe or \(5x^2 - 12x = 32\) or eg \(5y^2 - 6y - 155 (= 0)\) or \(5y^2 - 6y = 155\)	M1ft
\((5x + 8)(x - 4) (= 0)\) or \((x =)\dfrac{12 \pm \sqrt{(-12)^2 - 4 \times 5 \times (-32)}}{2 \times 5}\) or \(5\left[\left(x - \dfrac{6}{5}\right)^2 - \left(\dfrac{6}{5}\right)^2\right] - 32 (= 0)\) (should give \((x =) -\dfrac{8}{5}, 4\)) or eg \((5y - 31)(y + 5) (= 0)\) or \(\dfrac{6 \pm \sqrt{(-6)^2 - 4 \times 5 \times (-155)}}{2 \times 5}\) or \(5\left[\left(y - \dfrac{3}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2\right] - 155 (= 0)\) (should give \((y =) \dfrac{31}{5}, -5\))	M1ft
eg \(2 \times 4 + y = 3\) and \(2 \times -\dfrac{8}{5} + y = 3\) or eg \(2x + \dfrac{31}{5} = 3\) and \(2x - 5 = 3\)	M1ft
Working required if the correct answers come from incorrectly using \(y = 2x - 3\) oe award M4A0. Answer: \(x = -\dfrac{8}{5}\), \(y = \dfrac{31}{5}\), \(x = 4\), \(y = -5\)	A1
	(5)
	(5 marks)

Notes

M1: substitution of \(y = \pm 2x \pm 3\) (or \(x = \dfrac{\pm y \pm 3}{2}\)) into \(x^2 + y^2 = 41\) to obtain an equation in \(x\) only (or \(y\) only)

M1ft: dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of \(ax^2 + bx + c\;(= 0)\) where at least 2 coefficients (\(a\) or \(b\) or \(c\)) are correct

M1ft: dep on M1 method to solve their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as eg \(\dfrac{12 \pm \sqrt{144 + 640}}{10}\) or \(\dfrac{6 \pm \sqrt{36 + 3100}}{10}\) or if factorising allow brackets which expanded give 2 out of 3 terms correct) or correct values for \(x\) or correct values for \(y\)

M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations or their rearranged equation used in the substitution
or for one correct pair of values

A1: oe dep on M2 for all 4 values (allow coordinates)

If they find the values of \(y\) but think they are the values of \(x\) then the maximum mark is 3

Higher November 2024 Paper 1 Q23

EdexcelCurrent spec5 marksSolving Quadratics

23 The curve C has equation \(y = x^2 - 8x - 9\)
The straight line L has equation \(y = k\) where \(k\) is an integer.

C and L intersect at the points \(A\) and \(B\)

The coordinates of point \(A\) are \((p, k)\)
The coordinates of point \(B\) are \((q, k)\)

Given that \(p - q = 14\)

find the value of \(k\)
Show clear algebraic working.

(5)

Mark scheme

Scheme	Marks
\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 2x - 8\) or \((x - 4)^2\ldots\ldots\) or \((x =)\;\dfrac{-1 + 9}{2}\)	M1
\(x = 4\) or \((4, -25)\) or \((4, \ldots)\) or \(\dfrac{p + q}{2} = 4\) oe	M1
\(q = -3\) or \(p = 11\)	A1
\((k =)\;(-3)^2 - 8(-3) - 9\) or \((k =)\;(11)^2 - 8(11) - 9\)	M1
Working required Answer: 24	A1
	(5)
	(5 marks)

Notes

M1: for using differentiation or completing the square or by symmetry

A1: dep on M2

Mark scheme (ALT 1)

Scheme	Marks
\(p^2 - 8p - 9 = k\) and \(q^2 - 8q - 9 = k\) or \((p - 9)(p + 1) = k\) and \((q - 9)(q + 1) = k\)	M1
\((q + 14)^2 - 8(q + 14) - 9 = q^2 - 8q - 9\) or \((p - 14)^2 - 8(p - 14) - 9 = p^2 - 8p - 9\) or \((q + 14 - 9)(q + 14 + 1) = (q - 9)(q + 1)\) or \((p - 9)(p + 1) = (p - 14 - 9)(p - 14 + 1)\) oe	M1
\(q = -3\) or \(p = 11\)	A1
\((k =)\;(-3)^2 - 8(-3) - 9\) or \((k =)\;(11)^2 - 8(11) - 9\)	M1
Working required Answer: 24	A1

Notes

M1: for a correct equation in one variable

A1: dep on M2

Mark scheme (ALT 2)

Scheme	Marks
\((q + 14)^2 - 8(q + 14) - 9\;(= k)\) or \((p - 14)^2 - 8(p - 14) - 9\;(= k)\)	M1
\((q + 14)^2 - 8(q + 14) - 9 = q^2 - 8q - 9\) or \((p - 14)^2 - 8(p - 14) - 9 = p^2 - 8p - 9\)	M1
\(q = -3\) or \(p = 11\)	A1
\((k =)\;(-3)^2 - 8(-3) - 9\) or \((k =)\;(11)^2 - 8(11) - 9\)	M1
Working required Answer: 24	A1

Notes

M1: for a correct equation in one variable

A1: dep on M2

Mark scheme (ALT 3)

Scheme	Marks
\(y = (x - 4)^2 - 16 - 9\) oe or \(k = (x - 4)^2 - 25\) oe or \(x^2 - 8x - 9 - k = 0\)	M1
\((x =)\;4 \pm \sqrt{k + 25}\) or \((x =)\;\dfrac{--8 \pm \sqrt{(-8)^2 - (4 \times 1 \times (-9 - k))}}{2 \times 1}\) oe or \((x =)\;\dfrac{8 \pm \sqrt{100 + 4k}}{2}\) oe	M1
\(4 + \sqrt{k + 25} - \left(4 - \sqrt{k + 25}\right) = 14\) oe or \(\dfrac{8 + \sqrt{100 + 4k}}{2} - \left(\dfrac{8 - \sqrt{100 + 4k}}{2}\right) = 14\) oe	A1
\(25 + k = \left(\dfrac{14}{2}\right)^2\) oe	M1
Working required Answer: 24	A1

Notes

A1: for a correct equation in one variable

A1: dep on M2