Solve Quadratic Equation for Single or Multiple Intersections with the x-axis

[undefinable]

Homework Statement

Find k such that the graph of y=9x^2 + 3kx + k
A) intersects the x-axis in one point only
B) intersects the x-axis in two points
C) does not intersect the x-axis

Homework Equations

D= b^2 − 4ac
y= ax^2+bx+c
quadratic equation

The Attempt at a Solution

I got A but using the discriminent b^2-4ac= 0 meaning there is one root. but I have no idea how to get B and C. The answer is

B) k < 0 or k >4
C) 0<k<4

 

[sutupidmath]

$$y=9x^{2}+3kx+k$$, so a=9, b=3k, c=k, now

A) $$D=b^{2}-4ac=0$$ so we will have two identical solutions, and the graph will touch the x-axis. Probbably you meant here to touch , because it does not really intersect.
B) D>0
C)D<0

SInce a=9>0 it means that the parabola will be opened upward. So when D<0, it means that the parabola does not intersect the x-axis at all, since the quadratic equation $$9x^{2}+3kx+k=0$$ does not have real roots, while when D>0, it means that there are two distinct roots of the quadratic equation.

Do u know how to go about it now?

 

[undefinable]

Still don't understand.

For A, by using b^2-4ac= 0 I got the answer that k=0 (which is correct)
For B, I still don't understand how to achieve this answer k < 0 or k >4. How can I work with my equations to conclude that the values of k must be k< 0 or k >4 for y=9x^2 + 3kx + k to intersect two places.

 

[sutupidmath]

well B)$$D=b^{2}-4ac>0 =>(3k)^2-4*(9)*(k)>0=>9k^{2}-36k>0$$ , do you know how to solve this quadratic inequality?

One method for doint it is like this

$$9k^{2}-36k>0=>k^{2}-4k>0=> k(k-4)>0$$ then this is greater then zero if:1) k>0, and k-4>0, or 2) k<0 and k-4<0

another method is to first graph the function $$y=9k^{2}-36k$$ and see at what intervals the function is positive, that is above x-axis.

 

[tiny-tim]

For A, by using b^2-4ac= 0 I got the answer that k=0 (which is correct)

No.

b^2-4ac = 9k^2 - 36k.

9k^2 - 36k = 0 is a quadratic equation, and it has two solutions.

You only found one solution (k = 0) - what is the other?

 

[undefinable]

^ Right, 0 and 4, lol

I'm getting the numbers but I'm not getting the greater than and less than signs.

Referring to $$9k^{2}-36k>0=>k^{2}-4k>0=> k(k-4)>0$$ I got that but then what I concluded from that is k>0 and K>4. HOw can I end up with K<0 ?

 

[Tedjn]

Try graphing the quadratic y = k² - 4k. We are looking for when y > 0. What values of k give that?

 

[tiny-tim]

Right, 0 and 4, lol

Hurrah!

Referring to $$9k^{2}-36k>0=>k^{2}-4k>0=> k(k-4)>0$$ I got that but then what I concluded from that is k>0 and K>4. HOw can I end up with K<0 ?

Stop! … think! … k(k-4)>0 … try a few example for k …

Got it?

 

[sutupidmath]

^ Right, 0 and 4, lol

I'm getting the numbers but I'm not getting the greater than and less than signs.

Referring to $$9k^{2}-36k>0=>k^{2}-4k>0=> k(k-4)>0$$ I got that but then what I concluded from that is k>0 and K>4. HOw can I end up with K<0 ?

i perfectly well explained it in my post #4, that

$$k(k-4)>0$$ is possible in two cases


$$k>0 \ \ and \ \ k-4>0 \ or \ \ k<0 \ and \ \ k-4<0$$ think about this!

 

[Gib Z]

It may be easier to do k(k-4) > 0 graphically - the find the intervals for which the curve is above the k axis. Sketching that should be easy.