If alpha and beta are the zeros of polynomial 3 x^2-5x-9 find value of 2/ Alpha + 2/ beta.. Please solve this question fast, don’t post any irrelevant answer.. Class 10 maths About the author Harper
Answer: -10/9 Step-by-step explanation: I have already answered this question, you can check in my maths section recent answers. Reply
[tex]\large\underline{\sf{Solution-}}[/tex] Given that, [tex]\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {3x}^{2} – 5x – 9.[/tex] We know that, [tex]\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}[/tex] [tex]\bf\implies \: \alpha + \beta = – \dfrac{( – 5)}{3} = \dfrac{5}{3} – – – (1)[/tex] Also, [tex]\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}[/tex] [tex]\bf\implies \: \alpha \beta = \dfrac{ – 9}{ 3} = – 3 – – – (2)[/tex] Now, Consider, [tex]\rm :\longmapsto\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } [/tex] [tex]\rm \: \: = \: 2\bigg(\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \bigg) [/tex] [tex]\rm \: \: = \: 2\bigg(\dfrac{ \beta + \alpha }{ \alpha \beta } \bigg) [/tex] [tex]\rm \: \: = \: 2 \times \dfrac{( – 5)}{3} \times \dfrac{1}{3} [/tex] [tex]\rm \: \: = \: – \: \dfrac{10}{9} [/tex] [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } = – \: \dfrac{10}{9} }}[/tex] Additional Information :- [tex]\rm :\longmapsto\: \alpha,\beta \: and \: \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d[/tex] then, [tex]\green{\boxed{ \tt \: \alpha + \beta + \gamma = – \: \dfrac{b}{a} }}[/tex] [tex]\green{\boxed{ \tt \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}[/tex] [tex]\green{\boxed{ \tt \: \alpha \beta \gamma = – \: \dfrac{d}{a} }}[/tex] Reply
Answer:
-10/9
Step-by-step explanation:
I have already answered this question, you can check in my maths section recent answers.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {3x}^{2} – 5x – 9.[/tex]
We know that,
[tex]\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}[/tex]
[tex]\bf\implies \: \alpha + \beta = – \dfrac{( – 5)}{3} = \dfrac{5}{3} – – – (1)[/tex]
Also,
[tex]\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}[/tex]
[tex]\bf\implies \: \alpha \beta = \dfrac{ – 9}{ 3} = – 3 – – – (2)[/tex]
Now,
Consider,
[tex]\rm :\longmapsto\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } [/tex]
[tex]\rm \: \: = \: 2\bigg(\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \bigg) [/tex]
[tex]\rm \: \: = \: 2\bigg(\dfrac{ \beta + \alpha }{ \alpha \beta } \bigg) [/tex]
[tex]\rm \: \: = \: 2 \times \dfrac{( – 5)}{3} \times \dfrac{1}{3} [/tex]
[tex]\rm \: \: = \: – \: \dfrac{10}{9} [/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } = – \: \dfrac{10}{9} }}[/tex]
Additional Information :-
[tex]\rm :\longmapsto\: \alpha,\beta \: and \: \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d[/tex]
then,
[tex]\green{\boxed{ \tt \: \alpha + \beta + \gamma = – \: \dfrac{b}{a} }}[/tex]
[tex]\green{\boxed{ \tt \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}[/tex]
[tex]\green{\boxed{ \tt \: \alpha \beta \gamma = – \: \dfrac{d}{a} }}[/tex]