GCSE Maths revision · Higher: number and algebra
Quadratic simultaneous equations Higher only The idea With one linear and one quadratic equation, substitute the linear one into the quadratic. You get a quadratic in one letter, which usually gives two pairs of answers: the two crossing points.
Rearrange the linear equation: y = … y = \dots y = … or x = … x = \dots x = … . Substitute into the quadratic and solve. Put each value back into the linear equation to find its partner. Watch out Give both pairs, matched up correctly.
See it worked Worked example Calculator allowed · 3 marks Solve y = 2x + 1 and y = x 2 x^{2} x 2 − - − 2 simultaneously. Give both ordered pairs (x, y).
x 2 − 2 = 2 x + 1 x^{2}-2=2x+1 x 2 − 2 = 2 x + 1 ( x − 3 ) ( x + 1 ) = 0 (x-3)(x+1)=0 ( x − 3 ) ( x + 1 ) = 0 Equate the expressions for y: 2x + 1 = x 2 x^{2} x 2 − - − 2. Rearrange and factorise: x 2 x^{2} x 2 − - − 2x − - − 3 = (x − - − 3)(x + 1) = 0. The x-values are 3 and − 1 -1 − 1 ; substituting into y = 2x + 1 gives y = 7 and − 1 -1 − 1 respectively. Answer {(− 1, − 1), (3, 7)}
Your turn Question 1 Non-calculator · 6 marks The curve y = x 2 y=x^2 y = x 2 meets the line y = 6 x − 8 y=6x-8 y = 6 x − 8 .
(a) Find both intersection points exactly.
(b) Find the gradient of the chord joining the intersections.
Write your answer, then show the working (a) Equate the two expressions for y. x 2 = 6 x − 8 x^{2}=6x-8 x 2 = 6 x − 8 Factorise to obtain both x values. ( x − 2 ) ( x − 4 ) = 0 (x-2)(x-4)=0 ( x − 2 ) ( x − 4 ) = 0 Use the original curve to recover the matching y values. y = 2 2 y=2^{2} y = 2 2 Therefore ( 2 , 4 ) (2,4) ( 2 , 4 ) and ( 4 , 16 ) (4,16) ( 4 , 16 ) . (b) Use rise over run from the two points. 16 − 4 4 − 2 \frac{16-4}{4-2} 4 − 2 16 − 4 Therefore 6 6 6 . Answer (a) ( 2 , 4 ) (2,4) ( 2 , 4 ) and ( 4 , 16 ) (4,16) ( 4 , 16 ) ; (b) 6 6 6
Question 2 Non-calculator · 4 marks The line y = 2 x + k y = 2x + k y = 2 x + k touches the curve y = x 2 + 3 y = x^2 + 3 y = x 2 + 3 at exactly one point.
(a) Find the value of k k k .
(b) Write down the coordinates of the point where the line touches the curve.
Question 3 Non-calculator · 3 marks The line y = kx − - − 2 meets the curve y = x 2 x^{2} x 2 − - − 4x + 7 at exactly one point. Find both possible values of k.
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