# If the sum of the roots of a quadratic equation is 2 and the sum of their cubes is 98, then the equation is ?

2
by sowmyasreemajj

2014-10-05T13:20:52+05:30
So let the roots be
a,b

so given
a + b = 2
and
a³ + b³ = 98
⇒ (a + b)³ - 3ab(a + b) = 98
⇒ 2³ - 3ab × 2 = 98
⇒ -6ab = 90
⇒ ab = - 15

so the required equation =

x² - (a + b)x + ab = 0

⇒ x² - 2x - 15 = 0       ﻿﻿﻿ANSWER

hope it helps
plz mark as best
thank you
2014-10-05T13:22:57+05:30

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Let the roots of quadratic equations are m and n
sum of the roots = 2 : m+n=2
sum of their cubes = 98 : m³+n³=98

(m+n)³ = m³ + n³ + 3mn × (m+n)
⇒2³ = 98 + 3mn(2)
⇒ 8 = 98 + 6 mn
⇒ 6mn = 8-98 =-90
⇒ mn = -90/6
⇒ mn = -15

We know that for a quadratic equation x²+ax+b = 0,
sum of roots (m+n)= -b/1 = 2  ⇒ a = -2
product of roots(mn) = c/1 = -15  ⇒ b = -15

So equation is x² - 2x -15 = 0