Question

if alpha and bita are the zeroes if polynomial f(x)=ax2+bx+c then find 1/alpha 2+ 1/ bita2

Answer 1

Ax² + bx +c =0
zeroes α,β

sum of roots = α+β = -b/a
products of roots = αβ = c/a

1/α - 1/β = β-α / αβ   ............(1)

β-a = √(b² - 4ac/a² )      putting this value

1/α - 1/β  = √(b²-4ac/a²)  / c/a   =  √b²-4ac / c

hope this helps


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Answer 2

Answer:

      $\frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$  = $\frac{b^{2} - 2ac }{2c^{2} }$

Step-by-step explanation:

α and β are two zeroes of the polynomials that means α, β are two roots.

The given polynomial is: f(x) = ax² + bx + c

This is a quadratic polynomial.

• The roots will be,     α = $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$  and β = $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

• α² = ( $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$)² = $\frac{b^{2}+b^{2}-4ac - 2b\sqrt{b^{2}-4ac} }{4a^{2} }$ = $\frac{2b^{2}-4ac - 2b\sqrt{b^{2}-4ac} }{4a^{2} }$

• β² = ( $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$)² = $\frac{b^{2}+b^{2}-4ac + 2b\sqrt{b^{2}-4ac} }{4a^{2} }$ = $\frac{2b^{2}-4ac + 2b\sqrt{b^{2}-4ac} }{4a^{2} }$

• α² + β² =  $\frac{2b^{2}-4ac - 2b\sqrt{b^{2}-4ac} }{4a^{2} }$ +  $\frac{2b^{2}-4ac + 2b\sqrt{b^{2}-4ac} }{4a^{2} }$ = $\frac{b^{2} - 2ac }{2a^{2} }$

• αβ =   $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$ × $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$ = $\frac{-b^{2}+b^{2} - 4ac }{4a^{2} }$ = -c/a
• (αβ)² = (-c/a)² = c²/a²
• We have to find out, $\frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$

                     $\frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$ = $\frac{\alpha^{2} + \beta^{2} }{\alpha^{2}* \beta^{2} }$

                                 = $\frac{b^{2} - 2ac }{2a^{2} }$ ÷ $\frac{c^{2} }{a^{2} }$  [ putting all the values]

                                 =  $\frac{b^{2} - 2ac }{2a^{2} }$ × $\frac{a^{2} }{c^{2} }$

[ at the time of multiplication reciprocal will be taken.]

                        $\frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$   = $\frac{b^{2} - 2ac }{2c^{2} }$