# Thread: value of k and number of root

1. ## value of k and number of root

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.

2. ## Re: value of k and number of root

As is you cannot solve for k.
To find k you'd need to know how many real roots the equation has and then use the discriminant as appropriate.

Alternatively you can find x in terms of k using your favourite method

3. ## Re: value of k and number of root

Originally Posted by rcs
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 ...

$b^2 - 4ac < 0$ no real roots

$b^2 - 4ac = 0$ one real root of multiplicity two

$b^2 - 4ac > 0$ two real roots

so, if you're looking for the value of $k$ that provides at least one real root ...

$k^2 - 4(4)(6) \ge 0$

solve for $k$

4. ## Re: value of k and number of root

thanks... it this correct sir: k > = +- 4 (sqrt 6), is this the root now sir?

5. ## Re: value of k and number of root

Almost: $|k| \geq 4\sqrt{6}$ is true.

It follows that $k \geq 4\sqrt6$ which is one of your solutions yet we need to look at the other root: $-k \geq 4\sqrt{6} \longrightarrow \ k \leq -4\sqrt6$

6. ## Re: value of k and number of root

this one is not so bad.

if we use the idea of comparing coefficients as has been shown you can think like this.

$(A_1x+B_1)(A_2x+B_2)=A_1A_2x^2+A_1B_2x+A_2B_1x+B_1 B_2$

now you have $4x^2+kx+6$ all positive terms. the coefficients are 4 and 6. which factor to $A_1A_2=4=2\cdot 2\cdot 1$ and $B_1B_2=6=3\cdot 2\cdot 1$. so you have 2 factors of each coeffient just like the general form. now just substitute all the inormation into the general factors.

$(A_1x+B_1)(A_2x+B_2) \rightarrow (2x+2)(2x+3) \rightarrow 4x^2+10x+6$

$k=10$

now that you know k solve your linear equations to find the roots.

7. ## Re: value of k and number of root

skoker... im kind of confuse... how can you relate k = 10 from the answer given by e^(i*pi) above? how is it possible.. i think they are not the same

thanks

8. ## Re: value of k and number of root

I think I get it right no?

k=10 roots= -1 , -3/2.

I'm not sure what kind of business they have gotten into up there...

9. ## Re: value of k and number of root

here is a extra think...

to check k I do this.
$b^2-4ac>0 \rightarrow 10^2-4\cdot 4\cdot \6>0 \rightarrow 4>0$ 2 real roots.

now if we do like before we get this.
$k^2-4ac>0 \rightarrow k^2-4\cdot 4 \cdot \6>0 \rightarrow k^2-96>0 \rightarrow k^2>96 \rightarrow k>\pm 4 \sqrt{6}$

now we check with k and decimals. $\pm 4 \sqrt{6}=\pm 9.7980 \;,\; k=10$
$10>9.7980 \;,\; 10>-9.7980$

the inequalities is only telling you what value k is greater then or less then not k itself.

10. ## Re: value of k and number of root

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 $(A_1x + B_1)(A_2x + B_2)$ becomes $(2x+2)(2x+3)$

11. ## Re: value of k and number of root

Originally Posted by e^(i*pi)
I'm confused as to how $(A_1x + B_1)(A_2x + B_2)$ becomes $(2x+2)(2x+3)$
I think you guys over think on this one.

for quadratic equations this is true. all partial products of factors.
$(A_1x+B_1)(A_2x+B_2)=A_1A_2x^2+A_1B_2x+A_2B_1x+B_1 B_2$

and we have the rule $ax^2+bx+c \;,\; a_1c_1+a_2c_2=b$ 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.

$A_1=2 , A_2=2 , B_1=2 , B_2=3$