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2.13.2 Quadratic Functions, SPM Practice (Paper 2)


Question 3:
If α and β are the roots of the quadratic equation 3x2 + 2x– 5 = 0, form the quadratic equations that have the following roots.
(a)  2 α  and  2 β (b)  ( α + 2 β )  and  ( β + 2 α )

Solution:
3x2 + 2x – 5 = 0
a = 3, b = 2, c = –5
The roots are α and β.
α + β = − b a = − 2 3 α β = c a = − 5 3


(a)
The new roots are  2 α and 2 β . Sum of new roots = 2 α + 2 β = 2 β + 2 α α β = 2 ( α + β ) α β = 2 ( − 2 3 ) − 5 3 = 4 5

Product of new roots = ( 2 α ) ( 2 β ) = 4 α β = 4 − 5 3 = − 12 5

Using the formula, x2– (sum of roots)x + product of roots = 0
The new quadratic equation is
x 2 − ( 4 5 ) x + ( − 12 5 ) = 0
5x2 – 4x– 12 = 0



(b)
The new roots are  ( α + 2 β ) and ( β + 2 α ) . Sum of new roots = ( α + 2 β ) + ( β + 2 α )
= α + β + ( 2 α + 2 β ) = α + β + 2 α + 2 β α β = α + β + 2 ( α + β ) α β = 4 5 + 2 ( 4 5 ) − 12 5 = 4 5 − 2 3 = 2 15
Product of new roots = ( α + 2 β ) ( β + 2 α ) = α β + 2 + 2 + 4 α β
= − 12 5 + 4 + 4 − 12 5 = − 12 5 + 4 − 5 3 = − 1 15

The new quadratic equation is
x 2 − ( 2 15 ) x + ( − 1 15 ) = 0
15x2 – 2x– 1 = 0


Question 4:
It is given α and β are the roots of the quadratic equation x (x – 3) = 2k – 4, where k is a constant.
 (a) Find the range of values of k if α≠β. (b) Given  α 2  and  β 2  are the roots of another quadratic equation  2 x 2 +tx−4=0, where t is a constant, find the value of t and of k.

Solution:
(a) x( x−3 )=2k−4 x 2 −3x+4−2k=0 a=1, b=−3, c=4−2k     b 2 −4ac>0 ( −3 ) 2 −4( 1 )( 4−2k )>0    9−16+8k>0 8k>7   k> 7 8

(b) From the equation  x 2 −3x+4−2k=0, α+β=− b a  =− −3 1  =3………….( 1 ) αβ= c a = 4−2k 1 =4−2k………….( 2 ) From the equation 2 x 2 +tx−4=0, α 2 + β 2 =− t 2 α+β=−t………….( 3 ) α 2 × β 2 =− 4 2 αβ=−8………….( 4 ) Substitute (1)=(3), 3=−t t=−3 Substitute (2)=(4), 4−2k=−8 4+8=2k k=6

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