how is k solved when 4x^2 + kx + 6 = 0...
i really get so hard to solve when there is already k in the equation.
kindly help me.
Thanks a lot so much.
what the discriminant tells you ...
$\displaystyle b^2 - 4ac < 0$ no real roots
$\displaystyle b^2 - 4ac = 0$ one real root of multiplicity two
$\displaystyle b^2 - 4ac > 0$ two real roots
so, if you're looking for the value of $\displaystyle k$ that provides at least one real root ...
$\displaystyle k^2 - 4(4)(6) \ge 0$
solve for $\displaystyle k$
Almost: $\displaystyle |k| \geq 4\sqrt{6}$ is true.
It follows that $\displaystyle k \geq 4\sqrt6$ which is one of your solutions yet we need to look at the other root: $\displaystyle -k \geq 4\sqrt{6} \longrightarrow \ k \leq -4\sqrt6$
this one is not so bad.
if we use the idea of comparing coefficients as has been shown you can think like this.
$\displaystyle (A_1x+B_1)(A_2x+B_2)=A_1A_2x^2+A_1B_2x+A_2B_1x+B_1 B_2$
now you have $\displaystyle 4x^2+kx+6$ all positive terms. the coefficients are 4 and 6. which factor to $\displaystyle A_1A_2=4=2\cdot 2\cdot 1$ and $\displaystyle B_1B_2=6=3\cdot 2\cdot 1$. so you have 2 factors of each coeffient just like the general form. now just substitute all the inormation into the general factors.
$\displaystyle (A_1x+B_1)(A_2x+B_2) \rightarrow (2x+2)(2x+3) \rightarrow 4x^2+10x+6$
$\displaystyle k=10$
now that you know k solve your linear equations to find the roots.
here is a extra think...
to check k I do this.
$\displaystyle b^2-4ac>0 \rightarrow 10^2-4\cdot 4\cdot \6>0 \rightarrow 4>0$ 2 real roots.
now if we do like before we get this.
$\displaystyle k^2-4ac>0 \rightarrow k^2-4\cdot 4 \cdot \6>0 \rightarrow k^2-96>0 \rightarrow k^2>96 \rightarrow k>\pm 4 \sqrt{6}$
now we check with k and decimals. $\displaystyle \pm 4 \sqrt{6}=\pm 9.7980 \;,\; k=10$
$\displaystyle 10>9.7980 \;,\; 10>-9.7980$
the inequalities is only telling you what value k is greater then or less then not k itself.
To clarify mine and Skeeter's answers were stating the values of k for which there would be at least one real root.
Skoker has given the actual value of k. For what it's worth I'm confused as to how $\displaystyle (A_1x + B_1)(A_2x + B_2)$ becomes $\displaystyle (2x+2)(2x+3)$
I think you guys over think on this one.
for quadratic equations this is true. all partial products of factors.
$\displaystyle (A_1x+B_1)(A_2x+B_2)=A_1A_2x^2+A_1B_2x+A_2B_1x+B_1 B_2$
and we have the rule $\displaystyle ax^2+bx+c \;,\; a_1c_1+a_2c_2=b$ in this case a and c only have 2 factors each so they fall nicely into the LHS but you could use the same method for any ac coefficients that factor.
$\displaystyle A_1=2 , A_2=2 , B_1=2 , B_2=3$