If f(a) = a^2 + a + 1,then the number of solu | vedclass

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(C) Given $f(a) = a^2 + a + 1$.We need to solve $f(a^2) = 3f(a)$.Substituting $a^2$ into the function,we get $f(a^2) = (a^2)^2 + a^2 + 1 = a^4 + a^2 +

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$2$

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(C) Given $f(a) = a^2 + a + 1$.We need to solve $f(a^2) = 3f(a)$.Substituting $a^2$ into the function,we get $f(a^2) = (a^2)^2 + a^2 + 1 = a^4 + a^2 + 1$.The equation becomes $a^4 + a^2 + 1 = 3(a^2 + a + 1)$.$a^4 + a^2 + 1 = 3a^2 + 3a + 3$.$a^4 - 2a^2 - 3a - 2 = 0$.We can test for roots. If $a = -1$,$(-1)^4 - 2(-1)^2 - 3(-1) - 2 = 1 - 2 + 3 - 2 = 0$. So $(a+1)$ is a factor.Dividing $a^4 - 2a^2 - 3a - 2$ by $(a+1)$ gives $(a+1)(a^3 - a^2 - a - 2) = 0$.Testing $a=2$ in $a^3 - a^2 - a - 2$: $8 - 4 - 2 - 2 = 0$. So $(a-2)$ is a factor.Dividing $a^3 - a^2 - a - 2$ by $(a-2)$ gives $(a-2)(a^2 + a + 1) = 0$.Thus,the equation is $(a+1)(a-2)(a^2 + a + 1) = 0$.The roots are $a = -1$,$a = 2$,and the roots of $a^2 + a + 1 = 0$.The discriminant of $a^2 + a + 1$ is $D = 1^2 - 4(1)(1) = -3 Therefore,there are $2$ real solutions.

If $f(a) = a^2 + a + 1$,then the number of solutions of the equation $f(a^2) = 3f(a)$ is:

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