Quadratics

[cluelessinmath]

I really don't understand quadratics. I am trying to study for a big test but I still don't understand quadratics. Here is a problem from my homework that I got wrong:

Find the value of k that will make the solutions for the following quadractic equation.
ky ^2 - 4y + 1 =0

Select the correct answer:
a. k=7
b. k=4
c. k=1
d. k=2

So far i have done these steps:


x= -b +or- the square root of b^2 - 4ac
all over ____________
2a

(quadratic formula)

-(-4) + or - the sqare root of -4^2 - 2 (k) (1)
all over _______________
2(k)


4 + or - the square root of 16 -2k
all over _____________
2k

am I working the problem right? if so, were do I go from here? if not, where do i need to start? please help me!!!!!!!!!

 

[soroban]

Hello, cluelessinmath!

I really don't understand quadratics.

You say that ... but you're okay with the Quadratic Formula . . . good!

Find the value of k that will make the solutions . . make them what? ... equal?
for the following quadractic equation: ky² - 4y + 1 = 0

Select the correct answer:
a. k = 7 . . b. k=4 . . c. k=1 . . d. k=2

So far i have done these steps:
. . . . . . . . . . . . . _______
. . . . . . . . - b ± √b² - 4ac
. . . x . = . ------------------ . . . (Quadratic Formula)
. . . . . . . . . . . . .2a

. . . . . . . . . . . . . . .____________
. . . . . . . . -(-4) ± √(-4)² - 4(k)(1)
. . . y . = . ----------------------------
. . . . . . . . . . . . . . . 2(k)

. . . . . . . . . . . . .______
. . . . . . . . .4 ± √16 - 4k
. . . y . = . ----------------
. . . . . . . . . . . . . 2k


Am I working the problem right? . . . You're doing fine!
If so, where do I go from here?

If the solutions are to be <u>equal</u>, the discriminant (under the radical) must be zero.

So we have: . 16 - 4k .= .0 . . ---> . . k = 4 . . . answer (b)

 

[stapel]

You're on the right track: the quadratic formula is definitely the way to go (and it's the intended solution method, so you should get partial points, even if you make a mistake). Let's use single-line formatting, though, so our meaning is clear:

. . . . .x = [-b ± sqrt(b^2 - 4ac)] / 2a

. . . . .x = [-(-4) ± sqrt((-4)^2 - 4(k)(1))] / 2(k)

. . . . .x = [4 ± sqrt(16 - 4k)] / 2k

. . . . .x = [4 ± sqrt(4(4 - k))] / 2k

. . . . .x = [4 ± 2sqrt(4 - k)] / 2k

. . . . .x = [2 ± sqrt(4 - k)] / k

Obviously, k cannot equal zero, but you already knew that. The real point, however, is that, in order to have "solutions", you can't have a negative inside the square root. So all you really care about is that 4 - k isn't negative. (It can be zero, though.) So you need to solve:

. . . . .4 - k > 0

Eliz.