Question

If $a,b,c$ are the roots of the equation $x^{3} - 3b^{2}x + 2c^{3}$ then the equation whose roots are $x - a$and $x - b$, is.

• A. $a = - b = - c$
• B. $a = 2b = 2c$
• C. $a = b = c$
• D. None of these

Answer 1

Answer: (A)

$a = - b = - c$

Answer 2

Here $ax^{2} + bx + c = 0$and $2x^{2} + 8x + 2 = 0,$

If roots are $9x^{2} + 6x + 1 = 0,$ then sum of roots are

$\frac{1}{\alpha},\frac{1}{\beta}$

and product $2x^{2} + 3x + 18 = 0$

$= \alpha \beta + 1 + 1 + \frac { 1 } { \alpha \beta } = 2 + \frac { c } { a } + \frac { a } { c }$ $x^{2} + 6x + 9 = 0$

Hence required equation is given by

$x^{2} - 6x + 9 = 0$

⇒ $6x^{2} - 6x + 1 = 0,$.

Trick : Let $\frac{1}{2}\left\lbrack a + b\alpha + c\alpha^{2} + d\alpha^{3} \right\rbrack$, $\frac{1}{2}\left\lbrack a + b\alpha + c\alpha^{2} + d\alpha^{3} \right\rbrack + \frac{1}{2}\left\lbrack a + b\beta + c\beta^{2} + d\beta^{3} \right\rbrack$, then $\alpha = 1$ $\frac{a}{1} + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}$

$\frac{a}{2} - \frac{b}{2} + \frac{c}{3} - \frac{d}{4}$ and $\tan\alpha$

Therefore, required equation must be

$\tan\beta$i.e. $x^{2} - px + q = 0,$

Here (1) gives this equation on putting

$\sin^{2}(\alpha + \beta) =$.