Question

If
$\alpha \: and \: \beta$
are the roots of the quadratic equation x²+x-2=0, Find the value of
${ \alpha }^{ - 1 } - { \beta }^{ - 1}$


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Answer 1

$\bf \huge \bigstar \red{Answer : } \\ \\ \bf \: {x}^{2} + x - 2 \\ \\ \bf wkt : - \\ \\ a = 1 \\ b = 1 \\ c = - 2 \\ \\ \bf \: now \\ \\ \sf \alpha + \beta = \frac{ - b}{a} = - 1 \\ \\ \sf \alpha \beta = \frac{c}{a} = - 1 \\ \\ \bf \: then \\ \\ \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ \beta - \alpha }{ \alpha \beta } \\ \\ \implies \frac{ \sqrt{ {( \alpha + \beta )}^{2} - 4( - 2)} }{ \alpha \beta } \\ \\ \implies \: \frac{ \sqrt{ {( - 1)}^{2} + 8 } }{ - 2} \\ \\ \implies \: \frac{ \sqrt{9} }{ - 2} \\ \\ \implies \: \frac{ - 3}{2} \\ \\ \bf \boxed{value \: \: of \: \: \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ - 3}{2} }$

Answer 2

Step-by-step explanation:

★Answer:

x

2

+x−2

wkt:−

a=1

b=1

c=−2

now

α+β=

a

−b

=−1

αβ=

a

c

=−1

then

α

1

−

β

1

=

αβ

β−α

⟹

αβ

(α+β)

2

−4(−2)

⟹

−2

(−1)

2

+8

⟹

−2

9

⟹

2

−3

valueof

α

1

−

β

1

=

2

−3