Question

If alpha and beta are the roots of the equation ax2+bx+c=0 then the values of 1/alpha^2 + 1/beta^2 is

Answer 1

Given,

α and β are the roots of the equation ax²+bx+c = 0

To Find,

The value of 1/α²+1/β²

Solution,

Since α and β are the roots of the equation ax²+bx+c = 0

So,

Sum of roots = α+β = -b/a, where b is the coefficient of x and a is the coefficient of x²

Product of roots = αβ = c/a, where c is the constant.

Now,

1/α²+1/β² = (α²+β²)/α²β²

               = ((α+β)²-2αβ)/(c/a)²

               = ((-b/a)²-2(c/a))/c²/a²

               = (b²/a²-2c/a)/c²/a²

               = (b²-2ac)/a²*a²/c²

               = (b²-2ac)/c²

Hnece the value of 1/α²+1/β² is (b²-2ac)/c².

Answer 2

The value of $\frac{1}{\alpha ^{2} } +\frac{1}{\beta ^{2} }$   is  $\frac{b^{2}-2ac }{c^{2} }$ .

Step-by-step explanation:

Given:

$\alpha$ and $\beta$ are the roots of the $ax^{2} +bx+c=0.$

To Find:

The values of  $\frac{1}{\alpha ^{2} } +\frac{1}{\beta ^{2} }$ .

Formula Used:

If $\alpha$ and $\beta$ are the roots of the original quadratic $ax^{2} +bx+c=0$ or$x^{2} +\frac{b}{a} x+\frac{c}{a} =0.$ ---------- equation no.01.

Then  $(x-\alpha )(x-\beta )=0$

         $x^{2} -\alpha x-\beta x+\alpha \beta =0$

         $x^{2} - (\alpha +\beta) x+\alpha \beta =0$    ------ equation no.02.  

 But, A and B are the same equation.  

       $\alpha +\beta =\frac{-b}{a}$     and  $\alpha \beta =\frac{c}{a}$

Solution:

As given- $\alpha$ and $\beta$ are the roots of the $ax^{2} +bx+c=0.$

$\alpha +\beta =\frac{-b}{a}$     and  $\alpha \beta =\frac{c}{a}$

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

              $=\frac{(\alpha+\beta )^{2}-2\alpha \beta }{(\alpha \beta )^{2} }$

Putting value of  $\alpha +\beta =\frac{-b}{a}$     and  $\alpha \beta =\frac{c}{a}$.

              $=\frac{(\frac{-b}{a})^{2}-2(\frac{c}{a} ) }{(\frac{c}{a} )^{2} }$

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

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

Thus, the value of $\frac{1}{\alpha ^{2} } +\frac{1}{\beta ^{2} }$   is  $\frac{b^{2}-2ac }{c^{2} }$ .