Determine values of k for which function has two zeros.

**shadow6** · April 2nd 2010, 08:48 AM

Determine the values of k for which the function $f(x) = 4x^2 - 3x + 2kx +1$ has two zeros. Check the values in the original equation.

$a = 4, b= (-3 + 2k) , c = 1$
$b^2 - 4ac > 0$
$(-3 + 2k)^2 - 4(4)(1) > 0$
$9 + 4k^2 - 16 > 0$
$4k^2 > -9 + 16$
$\frac{4k^2 > 7}{4}$
$k^2 > \frac{7}{4}$
$\sqrt{k^2} > \sqrt{\frac{7}{4}}$
$k > \pm \frac{\sqrt{7}}{2}$

The textbook answer to this question is " $x < -0.5$ or $x > 3.5$ ". Where did I go wrong in solving this? Was it the part where I let $b = (-3 + 2k)$ ? I wasn't sure about that as there were two values of x in the standard form.

**alexmahone** · April 2nd 2010, 09:04 AM

$a=4, b=(-3 + 2k), c=1$

$b^2-4ac>0$

$(-3+2k)^2-4(4)(1)>0$

$9 + 4k^2-12k-16 > 0$

$4k^2-12k-7>0$

$4k^2+2k-14k-7>0$

$2k(2k+1)-7(2k+1)>0$

$(2k+1)(2k-7)>0$

$x<-0.5$ or $x>3.5$

**shadow6** · April 2nd 2010, 09:18 AM

Tyvm for the reply. Didn't even realize that I squared the bracketed numbers wrong. Just one question though. How did you get $(+2k - 14k)$ from $-12k$ ?

**jayAndy** · April 2nd 2010, 10:29 AM

Originally Posted by shadow6

Tyvm for the reply. Didn't even realize that I squared the bracketed numbers wrong. Just one question though. How did you get $(+2k - 14k)$ from $-12k$ ?

since -12k = 2k - 14k, it's like thinking backwards, he was able to plug it into the formula

he used this method to get an "extra" k to be able to factor the equation easier.

**CaptainBlack** · April 2nd 2010, 10:51 PM

All the posts in this thread present just a jumble of algebra, you will lose marks doing your work like that. Try explaining (in word, you remember them?) what you are doing at each stage. That way even if you make a mistake in the algebra and/or arithmetic you will still be credited with most of the available marks.

CB

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