0
$\begingroup$

I need help with the following.

The problem is stated like so: "The value of the constant $a$ is such that the quadratic function $f(x) \equiv x^2 +4x + a +3$ is never negative. Determine the nature of the roots of the equation $af(x) = (x^2+2)(a-1)$. Deduce the value of $a$ for which this equation has equal roots".

First statement implies that $b^2-4ac<0$, i.e. $a>1$

Now this part is strange. This is what you get (I hope) when you rewrite that second function (not sure why they call it $f(x)$ as well) :$f(x) = (a-1)x^2 + (2a-2)$. $b^2-4ac = -4(a-1)(2a-2)$, now since $a>1$ (not sure I can say this, because the functions are different; but since they denote this constant value as $a$ in both of these, I did consider it to be the same), the roots are complex because that result is negative.

It is actually not necessary to rewrite it like so, it is clear that roots are $x^2 = -2$, i.e. they are complex. However the answer is that the roots are real... I don't understand how this is possible.

I found that roots are equal when $a=1$, which is correct, according to the book.

$\endgroup$
2
$\begingroup$

Hint:

It seems to me that your interpretation of the second equation is wrong.

You have:

$$ a f(x)=(x^2+2)(a-1) \iff a(x^2+4x+a+3)=(x^2+2)(a-1) \mbox{ with } a\ge 1 $$ so you have to study the solutions of the equation $$ x^2+4ax+a^2+a+2=0 \mbox{ with } a \ge 1 $$

can you do this?

$\endgroup$
  • $\begingroup$ Ahhhh.... It's an equation, not an identity. I see now. Thx $\endgroup$ – i squared - Keep it Real Jul 25 '15 at 17:25
1
$\begingroup$

For the first part, you need to have $b^2-4ac\color{red}{\le} 0$, i.e. $a\ge 1$.

For the second part, we have $$a(x^2+4x+a+3)=(x^2+2)(a-1),$$ i.e. $$x^2+4ax+a^2+a+2=0.$$ Now the discriminant $$(4a)^2-4\cdot 1\cdot(a^2+a+2)=4(3a^2-a-2)=4(3a+2)(a-1)$$ is non-negative for $a\ge 1$.

For the third part, you are correct.

$\endgroup$
  • $\begingroup$ For the first question $a$ must be greater than $1,$ $(a>1).$ $\endgroup$ – Khadija Mbarki Jul 25 '15 at 17:15
  • 1
    $\begingroup$ @KhadijaMbarki: I don't think so. $f(x)$ can be zero. $\endgroup$ – mathlove Jul 25 '15 at 17:16
  • 1
    $\begingroup$ I'm with @mathlove. The phrase 'never negative' is the same as 'non-negative' i.e. $\geqslant 0$. $\endgroup$ – Kari Jul 25 '15 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.