AP EAPCET 05-Jul-2022 Shift 1 Paper

Let a,2a and b be roots of the cubic equation
x3+36−7x2=0
So, sum of roots =−‌

(−7)

1

∴‌a+2a+b=7
⇒3a+b=7
∴‌a+2a+b=7
⇒3a+b=7......(i)
Now, a(2a)b=36 [product of roots =‌

d

a

]
2a2b=36 and ab+a(2a)+2a(b)=0
⇒b=‌

−2a

3

⇒‌‌3a−‌

2a

3

=7
[from Eq. (i)]
⇒‌

9a−2a

3

=7⇒9a−2a=21
⇒7a=21⇒a=3
and b=7-3 a\;$
=7−3(3)=−2
[from Eq. (i)]
Roots are 3,6,−2
∴ Number of negative root is 1 .