Question

if f(a)=a^(2)+a+1,then number of solution of equation f(a^(2))=3f(a)

Answer 1

Number of solutions of a equation $f(a^{2}) = 3f(a)$ are two

Given : The quadratic equation f(a) = $a^{2}+a+1$

To Find : Number of solutions of a equation $f(a^{2}) = 3f(a)$

Solution : Number of solutions of a equation $f(a^{2}) = 3f(a)$ are two

Now a quadratic equation f(a) = $a^{2}+a+1$  is given

We have to find the number of solutions of a equation $f(a^{2}) = 3f(a)$

Now $f(a^{2})$ = $(a^{2}) ^{2} + a^{2}+1$

$f(a^{2})$ = $a^{4} + a^{2}+1$

$f(a^{2})$ = 3(f(a))

$f(a^{2})$ =  3( $a^{2}+a+1$ )

$a^{4} + a^{2}+1$ = $3a^{2}+3a+3$

g(a) = $a^{4} -2a^{2}-3a-2 =0$

Now g(-1) = 1 - 2 +3 -2

= -1 +1 = 0

Since g(-1) is zero

so -1 is one of the root of the equation

and g(2) = $2^{4 }- 2(2^{2}) -3(2)-2$

= 16 -8 -6-2

= 8-8 =0

So g(2) is zero

so 2 is also the root of the equation

So (a+1) and (a-2) is a factor of the equation g(a)

so $a^{2}-a-2$ will be the factor of g(a)

$a^{4} -2a^{2}-3a-2 = (a^{2}-a-2)(a^{2}+a+1)$

Here  $a^{2}+a+1$ does not have nay real solution

So 2 and -1 are the only roots

So number of solutions of a equation $f(a^{2}) = 3f(a)$ are two  (2 and -1)

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