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) 2α and 2β(b) (α+2β) and (β+2α)
Solution:
3x2 + 2x – 5 = 0
a = 3, b = 2, c = –5
The roots are α and β.
α+β=−ba=−23αβ=ca=−53
(a)
The new roots are 2αand2β.Sum of new roots=2α+2β=2β+2ααβ=2(α+β)αβ=2(−23)−53=45
Product of new roots=(2α)(2β)=4αβ=4−53=−125
Using the formula, x2– (sum of roots)x + product of roots = 0
The new quadratic equation is
x2−(45)x+(−125)=0
5x2 – 4x– 12 = 0
(b)
The new roots are (α+2β)and(β+2α).Sum of new roots=(α+2β)+(β+2α)
=α+β+(2α+2β)=α+β+2α+2βαβ=α+β+2(α+β)αβ=45+2(45)−125=45−23=215
Product of new roots=(α+2β)(β+2α)=αβ+2+2+4αβ
=−125+4+4−125=−125+4−53=−115
The new quadratic equation is
x2−(215)x+(−115)=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 α2 and β2 are the roots of another quadratic equation 2x2+tx−4=0, where t is a constant, find the value of t and of k.
Solution:
(a)x(x−3)=2k−4x2−3x+4−2k=0a=1, b=−3, c=4−2k b2−4ac>0(−3)2−4(1)(4−2k)>0 9−16+8k>08k>7 k>78
(b)From the equation x2−3x+4−2k=0,α+β=−ba =−−31 =3………….(1)αβ=ca=4−2k1=4−2k………….(2)From the equation 2x2+tx−4=0,α2+β2=−t2α+β=−t………….(3)α2×β2=−42αβ=−8………….(4)Substitute (1)=(3),3=−tt=−3Substitute (2)=(4),4−2k=−84+8=2kk=6
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 α2 and β2 are the roots of another quadratic equation 2x2+tx−4=0, where t is a constant, find the value of t and of k.
Solution:
(a)x(x−3)=2k−4x2−3x+4−2k=0a=1, b=−3, c=4−2k b2−4ac>0(−3)2−4(1)(4−2k)>0 9−16+8k>08k>7 k>78
(b)From the equation x2−3x+4−2k=0,α+β=−ba =−−31 =3………….(1)αβ=ca=4−2k1=4−2k………….(2)From the equation 2x2+tx−4=0,α2+β2=−t2α+β=−t………….(3)α2×β2=−42αβ=−8………….(4)Substitute (1)=(3),3=−tt=−3Substitute (2)=(4),4−2k=−84+8=2kk=6