Question

If ax^2 + bx + c = 0 has roots alpha and beta then thr quadratic equation cx^2 + bx + a =0 has roots​

Answer 1

Answer:

hope it helps you

Step-by-step explanation:

If ax

2

+bx+c=0 has roots α and β then-

aα

2

+bα+c=0 and aβ

2

+bβ+c=0

If α,β are the roots of x

2

−p(x+1)−c=0,c



=1 then, (α+1)(β+1)=1-c

Answer 2

ANSWER :

WE KNOW, x²-(sum of roots)x+product of roots=0.

So,

$\alpha + \beta = - \frac{b}{a} \underline{ \: \: \: \: \: \: \: }(1) \\ \alpha \beta = \frac{c}{a} \underline{ \: \: \: \: \: \: \: }(2) \: \: \: \: \: \: \: \: \: \:$

SOLUTION :

Let the roots are p and q

$c {x}^{2} + bx + a = 0$

MULTIPLYING -1/a BOTH SIDE

$\longrightarrow \: - \frac{c}{a} {x}^{2} - \frac{b}{a} x- 1 = 0$

$\longrightarrow \: - \alpha \beta {x}^{2} - (\alpha + \beta ) x- 1 = 0$

$\longrightarrow {x}^{2} + \frac{( \alpha + \beta )}{ \alpha \beta } x + \frac{1}{ \alpha \beta } = 0$

Hance,

$p + q= \frac{1}{ \alpha } + \frac{1}{ \beta }$

$\longrightarrow \: p = \frac{1}{ \alpha } + \frac{1}{ \beta } - q \underline{ \: \: \: \: \: \: \: \: }(3)$

and

$pq = \frac{1}{ \alpha \beta }$

$\longrightarrow \: p = \frac{1}{ \alpha \beta q} \underline{ \: \: \: \: \: \: \: }(4)$

From eqn three and four

$\frac{1}{ \alpha } + \frac{1}{ \beta } - q = \frac{1}{ \alpha \beta q}$

$\longrightarrow \: \beta q + \alpha q - {q}^{2} \alpha \beta = 0$

$\longrightarrow \: - {q}^{2} \alpha \beta + q( \alpha + \beta ) = 0$

$\longrightarrow \: q(q - \alpha - \beta ) = 0$

$\longrightarrow \boxed { \mathfrak{q = 0 \: \: \: or \: \: \: q = \alpha + \beta }}$

Put these two possible values of q in eqn 3

$\longrightarrow \boxed{ \mathfrak{p = { \frac{1}{ \alpha } + \frac{1}{ \beta } \: \: }}}{when \: q = 0}$

$\longrightarrow \boxed{ \mathfrak{p = { \frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha - \beta }}} \: \: \: \: when \: q = \alpha + \beta$