Question

Root of ax²+bx+c =o are
$\alpha \: and \: \beta$
find,
$1 \div \gamma + 1 \div \beta$
​

Answer 1

Solution

ax² + bx + c = 0

α and β are the roots

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

Taking LCM

$= \dfrac{1( \beta ) + 1( \alpha )}{ \alpha \beta }$

$= \dfrac{\beta + \alpha }{ \alpha \beta }$

$= \dfrac{ \alpha + \beta }{ \alpha \beta }$

By Quadratic formula

Roots of the equation :

• α = [ - b + √(b² - 4ac) ] / 2a
• β = [ - b - √(b² - 4ac) ] / 2a

Now, find the value of α + β and αβ

$\alpha + \beta = \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} + \dfrac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$= \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} + (- b - \sqrt{ {b}^{2} - 4ac})}{2a}$

$= \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} - b - \sqrt{ {b}^{2} - 4ac}}{2a}$

$= \dfrac{ - b }{a}$

i.e α + β = - b/a

$\alpha\beta = \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \times \dfrac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$= \dfrac{( - b + \sqrt{ {b}^{2} - 4ac} )( - b - \sqrt{ {b}^{2} - 4ac }) }{2a(2a)}$

$= \dfrac{( - b)^{2} - (\sqrt{ {b}^{2} - 4ac })^{2} }{4 {a}^{2} }$

$= \dfrac{b^{2} - ({b}^{2} - 4ac )}{4 {a}^{2} }$

$= \dfrac{b^{2} - {b}^{2} + 4ac}{4 {a}^{2} }$

$= \dfrac{ 4ac}{4 {a}^{2} }$

$= \dfrac{ c}{a }$

i.e αβ = c/a

Now, (α + β)/αβ

$= \dfrac{ - \frac{b}{a} }{ \frac{c}{a} }$

$= - \dfrac{b}{a} \times \dfrac{a}{c}$

$= - \dfrac{b}{c}$

i.e (α + β)/αβ = - b/c

Hence, the value if 1/α + 1/β is - b/c

Answer 2

Question :-

$</p><p> \sf{Root of \: \: \: a {x}^{2} + bx + c \: \: are } \\ \sf{</p><p>\alpha \: and \: \beta </p><p>find,} \: \frac{1}{ \alpha} + \frac{1}{ \beta}$

Answer :-

$\implies \: \boxed{\sf{ \frac{1}{ \alpha} + \frac{1}{ \beta} = \frac{ - b}{c} }} \:$

Formula used :-

$\star \boxed{ \sf{ sum \: of \: roots = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} }} } \\ \\ \star \boxed{\sf{product \: of \: roots = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}$

Explanation :-

Given Quadratic polynomial is

→ a x² + b x + c

$\sf{given \: that \: \alpha \: \: and \: \beta \: are \: roots \:} \\ \sf{of \: this \: eqation} \\ \\ \therefore \sf{ \alpha + \beta = \frac{ - b}{a} } \: \: \: \: \: ....eq.(1)\\ \sf{ and }\\ \rightarrow \: \sf{ \alpha \: \beta \: = \frac{c}{a} } \: \: \: \: ....eq.(2)$

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Now ,

$\rightarrow \: \frac{1}{ \alpha} + \frac{1}{ \beta} \\ \\ \rightarrow \: \frac{ \alpha \: + \: \beta}{ \alpha \: \beta \: } \\ \\ \sf{ from \: \: eq.(1) \: \: and \: eq.(2)} \\ \\ \rightarrow \: \sf{ \frac{ \frac{ - b}{a} }{ \frac{c}{a} } } \\ \\ \rightarrow \sf{ \frac{ - b}{c} } \\ \\ \therefore \boxed{\sf{ \frac{1}{ \alpha} + \frac{1}{ \beta} = \frac{ - b}{c} }}$

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