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Graph of two quadratic expressions $y=f(x)$ and $y=g(x)$ are shown in image, where $f(x) = x^2+2ax+b$ and $g(x) = cx^2+2dx+1$, (where $a$ $b$ $c$ $d$ are real)

Given that $|OB'| = 2|OA'|$ and $|BB'| = 2|AA'|$

If $|AA'| = 1$ and the equation $k^2(g(x))^2+(k-1)g(x)+2=0$ has exactly two real and distinct roots then the set of all possible values of k is.

graphs of f(x) and g(x)

I really don't have any idea on how should I proceed to solve this, any help would be appreciated.

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  • $\begingroup$ Is that an accurate representation of $f(x)$? Does it go through the origin like that? Or is that just an artifact of your sketch? $\endgroup$
    – scott
    Aug 30 '17 at 18:37
  • $\begingroup$ It does not go through the origin. $\endgroup$
    – Gem
    Aug 31 '17 at 1:35
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Hints:
1. If $|AA'|=1$ then the $y$-coordinate of the vertex of $f(x)$ is $-1$.
2. Complete the square on $f(x)$ and you will find the $x$-coordinate of the vertex is $-a$ and $b=(-a^2-1)$.
3. Clearly $c<0$.
4. Complete the square on $g(x)$ to find $\frac{d}{c}$, which should be twice $a$, given $|OB'| = 2|OA'|$.

Now you should have an idea what $c$ and $d$ are, so $k$ is your only variable. If that last equation has two real roots, it crosses the $x$-axis twice. Find the values of $k$ where this doesn't happen, and you have your range of answers.

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  • $\begingroup$ Shouldn't the x coordinate be -a $\endgroup$
    – Gem
    Aug 31 '17 at 1:38
  • $\begingroup$ @Gem you're right. Editing the answer. $\endgroup$
    – scott
    Aug 31 '17 at 15:43
  • $\begingroup$ b will also change accordingly $\endgroup$
    – Gem
    Aug 31 '17 at 15:46

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