Step-by-step explanation:
Solution :-
Given quadratic equation is
px²-6qx-(9p-10q) = 0
On comparing with the standard quadratic equation ax²+bx+c = 0
a = p
b = -6q
c = -(9p-10q) = -9p+10q
Given roots of the equation = 2 Ⴛ - 3 and
2 Ⴛ - 3 and 2β – 3
We know that
Sum of the roots = -b/a
=> 2 Ⴛ - 3 + 2β – 3 = -(-6q)/p
=> 2 Ⴛ + 2β - 6 = 6q/p
=> 2 Ⴛ + 2β = (6q/p)-6
=> 2 Ⴛ + 2β = (6q-6p)/p
=> 2(Ⴛ + β) = (6q-6p)/p
=> Ⴛ + β = (6q-6p)/(2p)
=> Ⴛ + β = 2(3q-3p)/(2p)
=> Ⴛ + β = (3q-3p)/p ---------(1)
Product of the roots = c/a
=> (2 Ⴛ - 3 )( 2β – 3) = (-9p+10q)/p
=> 4Ⴛ β-6Ⴛ - 6β+9 = (-9p+10q)/p
=> 4Ⴛ β-6(Ⴛ +β)+9 = (-9p+10q)/p
=> 4Ⴛ β-6[(3q-3p)/p]+9 = (-9p+10q)/p
=> 4Ⴛ β-6[(3q-3p)/p] =[ (-9p+10q)/p]-9
=> 4Ⴛ β-6[(3q-3p)/p] = (-9p+10q-9p)/p
=> 4Ⴛ β-6[(3q-3p)/p] = (-18p+10q)/p
=> 4Ⴛ β =[(-18p+10q)/p]+6[(3q-3p)/p]
=> 4Ⴛ β =[(-18p+10q)/p]+[(18q-18p)/p]
=> 4Ⴛ β =[(-18p+10q)+18q-18p)]/p
=> 4Ⴛ β = (-36p+28q)/p
=> 4Ⴛ β = 4(-9p+7q)/p
=> Ⴛ β = 4(-9p+7q)/4p
=> Ⴛ β = (-9p+7q)/p
=> Ⴛ β = (7q-9p)/p ----------(2)
We know that
The quadratic equation whose roots are Ⴛ and β is x²-(Ⴛ+ β)x +Ⴛ β = 0
=> x²- [ (3q-3p)/p]x + [ (7q-9p)/p] = 0
=> [px²-(3q-3p)x+(7q-9p)]/p = 0
=> [px²-(3q-3p)x+(7q-9p)] = 0×p
=> px²-(3q-3p)x+(7q-9p) = 0
Answer :-
The required quadratic equation is
px²-(3q-3p)x+(7q-9p) = 0
Used Formulae:-
- The standard quadratic equation is ax²+bx+c = 0
- The quadratic equation whose roots are Ⴛ and β is x²-(Ⴛ+ β)x +Ⴛ β = 0
- Sum of the roots = -b/a
- Product of the roots = c/a
