Question

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

Answer 1

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

Answer 2

Given Equation

$\tt \to \: {x}^{2} + 7x + 5 = 0$

To Find

$\tt\to \dfrac{1}{ \alpha ^{2} } + \dfrac{1}{ { \beta }^{2} }$

Now Compare with

$\tt \to {ax}^{2} + bx + c = 0$

We Get

$\tt \to \: a = 1,b = 7 \: and \: c = 5$

Sum of the root are

$\tt \to \: ( \alpha + \beta ) = \dfrac{ - b}{a}$

Product of the root are

$\tt \to \: ( \alpha \beta ) = \dfrac{c}{a}$

We get

$\tt \to \: ( \alpha + \beta ) = \dfrac{ - 7}{1} = - 7$

$\tt \to \: ( \alpha \beta ) = \dfrac{5}{1} = 5$

Now we Have to find

$\tt\to \dfrac{1}{ \alpha ^{2} } + \dfrac{1}{ { \beta }^{2} }$

Taking Lcm

$\tt \to \: \dfrac{ { \alpha }^{2} + { \beta }^{2} }{( \alpha \beta ) {}^{2} }$

$\tt \to \: \dfrac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{( \alpha \beta ) {}^{2} }$

$\tt \to \: \dfrac{( - 7) {}^{2} - 2 \times 5 }{(5) ^{2} }$

$\tt \to \: \dfrac{49 - 10}{25} = \dfrac{39}{25}$

Answer

$\tt \to \: \dfrac{39}{25}$