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If α,β are the roots of $x^{2} - (a - 2) x - (a + 1) = 0$ where a is a variable, then the least value of $\propto^{2} + β^{2}$ is:

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answer is C.

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Detailed Solution

The given equation is

x^{2} - (a - 2) x - (a + 1) = 0

and its roots are α and β.
We know that if α and β are the roots of the equation

a x^{2} + bx + c = 0

then α+β=−

\frac{b}{a}

and αβ=

\frac{c}{a}

.
Now, for the equation

x^{2} - (a - 2) x - (a + 1) = 0

we can have b=−(a−2) , c=−(a+1) and a=1.
Therefore, we can write α+β=−

\frac{b}{a}

=−

\frac{- (a - 2)}{1}

and αβ=

\frac{c}{a}

\frac{- (a - 2)}{1}

We know that

{(α + β)}^{2} = α^{2} + 2 \propto β + β^{2}

By manipulating the above equation. We got the equation.
⇒

α^{2} + β^{2}

{(α + β)}^{2}

−2αβ
Now, we can put the values α+β and αβ in the above equation.
⇒

α^{2} + β^{2}

{(- (a - 2))}^{2}

−2(−(a+1))
⇒

α^{2} + β^{2}

a^{2}

−4a+4+2a+2
Now, simplify the above equation
⇒

α^{2} + β^{2} = a^{2} - 2 a + 6

Now, the above equation can be written as
⇒

α^{2} + β^{2} = a^{2} - 2 a + 1 + 5

⇒

α^{2} + β^{2} = ({a - 1)}^{2} + 5

Now, from the above equation we can say that if the value of a is 1 then we will get the minimum value of

α^{2} + β^{2}

is
⇒

α^{2} + β^{2}

({1 - 1)}^{2} + 5

=5
Therefore, we got the least value

α^{2} + β^{2}

is 5
Hence, the correct option is (3)

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