Question

If for all real values of 'a' ,one of the zero of the polynomial x^2 -3ax+f(a) is double the other zero, then find f(x)

Answer 1

$roots \: are \: \alpha \: and \: 2 \alpha \\ \alpha + 2 \alpha = \frac{3a}{1} \\ 3 \alpha =3 a \\ \alpha = a \\ \alpha .2 \alpha = \frac{ f(a)}{1} \\ 2 { \alpha }^{2} = f(a) \\ 2 {a}^{2} = \: f(a)$

Answer 2

f(x) = 2x² If for all real values of 'a' ,one of the zero of the polynomial x² -3ax+f(a) is double the other zero

Step-by-step explanation:

Let say roots are

β  & 2β

Sum of roots = - (-3a)/1

=> β + 2β = 3a

=> 3β = 3a

=> β = a

Product of roots = f(a)/1

=> β * 2β = f(a)

=> a * 2a = f(a)

=> f(a) = 2a²

=> f(x) = 2x²

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