
If $ \alpha $ and $ \beta $ are zeros of the quadratic polynomial $ 2{x^2} + 3x - 6 $ , then find the values of $ {\alpha ^2} + {\beta ^2} $ .
Answer
510.9k+ views
Hint: General form of quadratic equation is $ a{x^2} + bx + c = 0 $
Roots of quadratic equation is given by formula $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Roots are also called zeros of the quadratic equation.
Complete step-by-step answer:
Given polynomial $ 2{x^2} + 3x - 6 $
Roots of the above polynomial (quadratic expression) is $ \alpha $ and $ \beta $ .
So, as in quadratic polynomial:
Sum of roots $ = \dfrac{{ - b}}{a} = \alpha + \beta $
Product of roots $ = \dfrac{c}{a} = \alpha \beta $
Now, in case of our polynomial:
i.e. $ 2{x^2} + 3x - 6 $ 2 on comparing with $ a{x^2} + bx + c $
Here, $ a = 2 $
$ b = 3 $
$ c = - 6 $
So,
Sum of roots $ = \dfrac{{ - b}}{a} = \dfrac{{ - 3}}{2} = \alpha + \beta $
Product of roots $ = \dfrac{c}{a} = \dfrac{{ - 6}}{2} = - 3 = \alpha \beta $
Now we have to find out the value of $ {\alpha ^2} + {\beta ^2} $
As we know, $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
$ \Rightarrow {a^2} + {b^2} = {(a + b)^2} - 2ab $
So,
Putting the value of \[a = \alpha \,,\,b = \beta \],
We get,
$ {\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta $ …..(i)
In our case, $ \alpha + \beta = \dfrac{{ - 3}}{2}\,\,\alpha \beta = - 3 $
So, putting $ (\alpha + \beta ) $ and $ \alpha \beta $ in equation (i)
We get,
$ {\alpha ^2} + {\beta ^2} = {\left( {\dfrac{{ - 3}}{2}} \right)^2} - 2( - 3) $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = \dfrac{9}{4} + 6 $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = \dfrac{{9 + 24}}{4} $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = \dfrac{{33}}{4} $
So, $ {\alpha ^2} + {\beta ^2} = \dfrac{{33}}{4} $
Note: For general quadratic expression for i.e. $ a{x^2} + bx + c $
Sum of roots $ = \dfrac{{ - b}}{a} $
Product of roots $ = \dfrac{c}{a} $
Roots of quadratic $ = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
We also need to be smart enough to use algebraic identities and manipulate them as per the requirements of what has been asked.
Roots of quadratic equation is given by formula $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Roots are also called zeros of the quadratic equation.
Complete step-by-step answer:
Given polynomial $ 2{x^2} + 3x - 6 $
Roots of the above polynomial (quadratic expression) is $ \alpha $ and $ \beta $ .
So, as in quadratic polynomial:
Sum of roots $ = \dfrac{{ - b}}{a} = \alpha + \beta $
Product of roots $ = \dfrac{c}{a} = \alpha \beta $
Now, in case of our polynomial:
i.e. $ 2{x^2} + 3x - 6 $ 2 on comparing with $ a{x^2} + bx + c $
Here, $ a = 2 $
$ b = 3 $
$ c = - 6 $
So,
Sum of roots $ = \dfrac{{ - b}}{a} = \dfrac{{ - 3}}{2} = \alpha + \beta $
Product of roots $ = \dfrac{c}{a} = \dfrac{{ - 6}}{2} = - 3 = \alpha \beta $
Now we have to find out the value of $ {\alpha ^2} + {\beta ^2} $
As we know, $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
$ \Rightarrow {a^2} + {b^2} = {(a + b)^2} - 2ab $
So,
Putting the value of \[a = \alpha \,,\,b = \beta \],
We get,
$ {\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta $ …..(i)
In our case, $ \alpha + \beta = \dfrac{{ - 3}}{2}\,\,\alpha \beta = - 3 $
So, putting $ (\alpha + \beta ) $ and $ \alpha \beta $ in equation (i)
We get,
$ {\alpha ^2} + {\beta ^2} = {\left( {\dfrac{{ - 3}}{2}} \right)^2} - 2( - 3) $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = \dfrac{9}{4} + 6 $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = \dfrac{{9 + 24}}{4} $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = \dfrac{{33}}{4} $
So, $ {\alpha ^2} + {\beta ^2} = \dfrac{{33}}{4} $
Note: For general quadratic expression for i.e. $ a{x^2} + bx + c $
Sum of roots $ = \dfrac{{ - b}}{a} $
Product of roots $ = \dfrac{c}{a} $
Roots of quadratic $ = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
We also need to be smart enough to use algebraic identities and manipulate them as per the requirements of what has been asked.
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