If the roots of the equation {x^3} - 12{x^2} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) Let $a -d, a, a + d$ be the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$

Then $(a - d) + a + (a + d) = 12$ and $(a - d)\,a\,(a + d) =

Vedclass · Accepted Answer

$ \pm 3$

Vedclass · Answer

(c) Let $a -d, a, a + d$ be the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$

Then $(a - d) + a + (a + d) = 12$ and $(a - d)\,a\,(a + d) = 28$

==> $3a = 12$ and $a\,({a^2} - {d^2}) = 28$

==> $a = 4$ and $a\,({a^2} - {d^2}) = 28$

==> $16 - {d^2} = 7$

$\Rightarrow d = \pm \,3$.

If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$, then their common difference will be

Similar Questions

If ${a^2},\,{b^2},\,{c^2}$ be in $A.P.$, then $\frac{a}{{b + c}},\,\frac{b}{{c + a}},\,\frac{c}{{a + b}}$ will be in

Three number are in $A.P.$ such that their sum is $18$ and sum of their squares is $158$. The greatest number among them is

If the roots of the equation $x^3 - 9x^2 + \alpha x - 15 = 0 $ are in $A.P.$, then $\alpha$ is

There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is

The difference between an integer and its cube is divisible by