If roots of equation x²+7x+5 = 0 are α & β. Find 1/α ²+ 1/β²​

2 thoughts on “If roots of equation x²+7x+5 = 0 are α & β. Find 1/α ²+ 1/β²​”

1. Given Equation

   [tex] \tt \to \: {x}^{2} + 7x + 5 = 0[/tex]

   To Find

   [tex] \tt\to \dfrac{1}{ \alpha ^{2} } + \dfrac{1}{ { \beta }^{2} } [/tex]

   Now Compare with

   [tex] \tt \to {ax}^{2} + bx + c = 0[/tex]

   We Get

   [tex] \tt \to \: a = 1,b = 7 \: and \: c = 5[/tex]

   Sum of the root are

   [tex] \tt \to \: ( \alpha + \beta ) = \dfrac{ – b}{a} [/tex]

   Product of the root are

   [tex] \tt \to \: ( \alpha \beta ) = \dfrac{c}{a} [/tex]

   We get

   [tex]\tt \to \: ( \alpha + \beta ) = \dfrac{ – 7}{1} = – 7[/tex]

   [tex] \tt \to \: ( \alpha \beta ) = \dfrac{5}{1} = 5[/tex]

   Now we Have to find

   [tex]\tt\to \dfrac{1}{ \alpha ^{2} } + \dfrac{1}{ { \beta }^{2} } [/tex]

   Taking Lcm

   [tex] \tt \to \: \dfrac{ { \alpha }^{2} + { \beta }^{2} }{( \alpha \beta ) {}^{2} } [/tex]

   [tex] \tt \to \: \dfrac{( \alpha + \beta ) {}^{2} – 2 \alpha \beta }{( \alpha \beta ) {}^{2} } [/tex]

   [tex] \tt \to \: \dfrac{( – 7) {}^{2} – 2 \times 5 }{(5) ^{2} } [/tex]

   [tex] \tt \to \: \dfrac{49 – 10}{25} = \dfrac{39}{25} [/tex]

   Answer

   [tex] \tt \to \: \dfrac{39}{25} [/tex]

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2. Answer:

   39/25

   Step-by-step explanation:

   For an equation in form of ax² + bx + c = 0, -b/a and c/a is the sum and product of roots respectively. Therefore, if α and β are roots:

   ⇒ -(7/1) = α + β and 5/1 = αβ

   ⇒ – 7 = α + β and 5 = αβ

   Square on both sides of α+β

   ⇒ 49 = α² + β² + 2αβ

   ⇒ 49 – 2(5) = α² + β² [αβ = 5]

   ⇒ 39 = α² + β²

   ∴ 1/α² + 1/β² = (β² + α²)/(αβ)²

   = 39/(5)²

   = 39/25

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