If a,b,c are real and {x^3} - 3{b^2}x + 2{c^3 | vedclass

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

Question

(c) As $f(x) = {x^3} - 3{b^2}x + 2{c^3}$ is divisible by $x - a$ and $x - b$, therefore
$f(a) = 0\,\,$

$\Rightarrow {a^3} - 3{b^2}a + 2{c^3} = 0$.

Vedclass · Accepted Answer

$a = b = c$,$a = - 2b = - 2c$

Vedclass · Answer

(c) As $f(x) = {x^3} - 3{b^2}x + 2{c^3}$ is divisible by $x - a$ and $x - b$, therefore
$f(a) = 0\,\,$

$\Rightarrow {a^3} - 3{b^2}a + 2{c^3} = 0$.....$(i)$

and $f(b) = 0$==>${b^3} - 3{b^3} + 2{c^3} = 0$.....$(ii)$

From $(ii),$ $b = c$

From $(i)$, ${a^3} - 3a{b^2} + 2{b^3} = 0$ (Putting $b = c$)

$ \Rightarrow (a - b)({a^2} + ab - 2{b^2}) = 0$

$ \Rightarrow $$a = b$ or ${a^2} + ab = 2b$

Thus $a = b = c$ or ${a^2} + ab = 2{b^2}$ and $b = c$

${a^2} + ab = 2{b^2}$is satisfied by $a = - 2b$. But $b = c$.

$	herefore$ ${a^2} + ab - 2{b^2}$ and $b = c$ is equivalent to $a = - 2b = - 2c$

If $a,b,c$ are real and ${x^3} - 3{b^2}x + 2{c^3}$ is divisible by $x - a$ and$x - b$, then

Similar Questions

The set of all $a \in R$ for which the equation $x | x -1|+| x +2|+a=0$ has exactly one real root is:

All the points $(x, y)$ in the plane satisfying the equation $x^2+2 x \sin (x y)+1=0$ lie on

The number of distinct real roots of the equation $|\mathrm{x}+1||\mathrm{x}+3|-4|\mathrm{x}+2|+5=0$, is ...........

The number of solution$(s)$ of the equation $2^x = x^2$ is

Let $p_1(x)=x^3-2020 x^2+b_1 x+c_1$ and $p_2(x)=x^3-2021 x^2+b_2 x+c_2$ be polynomials having two common roots $\alpha$ and $\beta$. Suppose there exist polynomials $q_1(x)$ and $q_2(x)$ such that $p_1(x) q_1(x)+p_2(x) q_2(x)=x^2-3 x+2$. Then the correct identity is