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the roots of the equation, px^2 - 6qx - (9p - 10q) = 0 are 2 ⴛ - 3 and 2β – 3, find the equation whose roots are ⴛ and β​

Sagot :

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
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