Question

if alpha and beta are the roots of ax^2+bx+c then find 1/alpha - 1/beta​

Answer 1

Question :

If α and β are the roots of ax² + bx + c = 0. Then, find 1/α - 1/β.

Solution:

We know that,

• α + β = -b/c
• αβ = c/a

Now,

$( \alpha - \beta )^{2} = \alpha ^{2} + \beta^{2} - 2 \alpha \beta \: \: \: \: \: \\ \\ ( \alpha - \beta )^{2} =( \alpha + \beta ) ^{2}- 4 \alpha \beta \: \: \: \: \\ \\ ( \alpha - \beta )^{2} = (\frac{ -b}{a}^{2} ) - \frac{4c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ( \alpha - \beta )^{2} = \frac{ b^{2} - 4c}{ {a}^{2} } \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \alpha - \beta = \sqrt{\frac{ b^{2} - 4c}{ {a}^{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \\ \\ \bold{\alpha - \beta = \pm \frac{ \sqrt{ b^{2} - 4c}}{ {a}^{2} }}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \:$

Now,

$\dfrac{1}{ \alpha } - \dfrac{1}{ \beta }$

$\implies \dfrac{ \alpha - \beta }{ \alpha \beta }$

$\implies \dfrac{ \pm \: \sqrt{{b}^{2} - 4ac}}{a} \times \dfrac{a}{c}$

$\boxed{ \bf \therefore \: \dfrac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ \pm \: \sqrt{{b}^{2} - 4ac}}{c }}$

Answer 2

Step-by-step explanation:

We know that,

α + β = -b/c

αβ = c/a

Now,

\begin{gathered}( \alpha - \beta )^{2} = \alpha ^{2} + \beta^{2} - 2 \alpha \beta \: \: \: \: \: \\ \\ ( \alpha - \beta )^{2} =( \alpha + \beta ) ^{2}- 4 \alpha \beta \: \: \: \: \\ \\ ( \alpha - \beta )^{2} = (\frac{ -b}{a}^{2} ) - \frac{4c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ( \alpha - \beta )^{2} = \frac{ b^{2} - 4c}{ {a}^{2} } \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \alpha - \beta = \sqrt{\frac{ b^{2} - 4c}{ {a}^{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \\ \\ \bold{\alpha - \beta = \pm \frac{ \sqrt{ b^{2} - 4c}}{ {a}^{2} }}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \end{gathered}

(α−β)

2

=α

2

+β

2

−2αβ

(α−β)

2

=(α+β)

2

−4αβ

(α−β)

2

=(

a

−b

2

)−

a

4c

(α−β)

2

=

a

2

b

2

−4c

α−β=

a

2

b

2

−4c

α−β=±

a

2

b

2

−4c

Now,

\dfrac{1}{ \alpha } - \dfrac{1}{ \beta }

α

1

−

β

1

\implies \dfrac{ \alpha - \beta }{ \alpha \beta }⟹

αβ

α−β