Question

If α and β are the zeroes of a quadratic polynomial x2 + x – 2 then find the value of

(i) 1/ α + 1/ β

(ii) α2 + β

​

Answer 1

Answer:

Discriminant (d) =

${b}^{2} - 4ac \: = \: {1}^{2} - 4 \times 1 \times ( - 2) \: 1 + 8 \: = 9$

$\alpha = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 1 + \sqrt{9} }{2 \times 1} = \frac{ - 1 + 3}{2} = \frac{2}{2} = 1$

$\beta = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 1 - \sqrt{9} }{2 \times 1} = \frac{ - 1 - 3}{2} = \frac{ - 4}{2} = - 2$

(i)

$\frac{1}{a} + \frac{1}{b} = \frac{1}{1} + \frac{1}{ - 2} = 1 - \frac{1}{2} = 1 + 0.5 = 1.5$

(ii)

${a}^{2} + {b}^{2} = {1}^{2} + { - 2}^{2} = 1 + 4 = 5$

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Answer 2

★${\underline{\huge{\mathbb{Question:-}}}}$

If α and β are the zeroes of a quadratic polynomial x2 + x – 2 then find the value of

(i) 1/ α + 1/ β

(ii) α²+ β²

★${\underline{\huge{\mathbb{Solution:-}}}}$

If α and β are the zeroes of a quadratic polynomial x²+ x – 2.

x²+x-2=0.....................(i)

•• Compare (i) no. equation with ax²+bx+c = 0.

So,

a = 1

b= 1

c=(-2)

${\red{\boxed{\bold{\alpha+\beta=\frac{-b}{a}=\frac{-1}{1}=-1}}}}$

${\red{\boxed{\bold{\alpha×\beta=\frac{c}{a}=\frac{-2}{1}=-2}}}}$

(i)$\frac{1}{\alpha}+\frac{1}{\beta}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } \\ = \frac{ \beta + \alpha }{ \alpha \beta } \\ = \frac{ - 1}{ - 2} \\ = \frac{1}{2}$

(ii)

${ \alpha }^{2} + { \beta }^{2} \\ = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ = {( - 1)}^{2} - 2 \times - 2 \\ = 1 + 4 \\ = 5$

★${\underline{\huge{\mathbb{Answer:-}}}}$

(I)$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{2}=0.5$

(ii)$\alpha^2+\beta^2=5$