# [SOLVED] High School Quadratic Inequalities

• October 1st 2009, 09:13 AM
unstopabl3
[SOLVED] High School Quadratic Inequalities
Q) Use the discriminant 'b^2=-4ac' to solve the following:

i) Find the values of ' k ' for which the following equations have two separate roots.

a) $kx^2+kx+2=0$

I solved the above to get

$k(k-8)>0$

Now my question is that we know that k>8 but how do we figure k<0?

Is it because we are 100% that if k>8 from my final step then the other value must be smaller? (k<0)

ii) Find the values of ' k ' for which the following equations have no roots.

a) $k^2x^2+2kx+1=0$

I solved this and got:

$4k^2-4k^2<0$

$0<0$

How is the answer k=0? I mean shouldn't 0<0 mean no solution or something?

iii) Sketch, on the same diagram, the graphs of $y=1/x$ and $y=x-3/2$. Find the solution set of the inequality $x-3/2>1/x$

Please tell me step by step on how to solve this question.

• October 1st 2009, 10:54 AM
Amer
Quote:

Originally Posted by unstopabl3
Q) Use the discriminant 'b^2=-4ac' to solve the following:

i) Find the values of ' k ' for which the following equations have two separate roots.

a) $kx^2+kx+2=0$

I solved the above to get

$k(k-8)>0$

Now my question is that we know that k>8 but how do we figure k<0?

Is it because we are 100% that if k>8 from my final step then the other value must be smaller? (k<0)

ii) Find the values of ' k ' for which the following equations have no roots.

a) $k^2x^2+2kx+1=0$

I solved this and got:

$4k^2-4k^2<0$

$0<0$

How is the answer k=0? I mean shouldn't 0<0 mean no solution or something?

iii) Sketch, on the same diagram, the graphs of $y=1/x$ and $y=x-3/2$. Find the solution set of the inequality $x-3/2>1/x$

Please tell me step by step on how to solve this question.

for the first question

k(k-8)>0 you want the values of k that make this inequality true just draw the real number line and plot 0 and 8 (zero of k(k-8)) then take a value between 0,8 and sub it in k(k-8) if the sign is negative ignore the interval between 0,8 take a number more than 8 and sub if it is positive take this interval and take a number less than 0 and sub if it is positive take the interval (-infinity , 0)

ii) if $b^2-4ac < 0$ there is no solution right but here

$b^2-4ac = 4k^2 - 4k^2 = 0$ for all k real number so always there is a equal roots and the root is

$\frac{-b \mp \sqrt{4k^2-4k^2}}{2a} = \frac{-b}{2a} = \frac{-2k}{2k^2}=\frac{-1}{k}$ repeated root so there is no value of k that make no solution

iii) draw the two curve and take the region where $x-\frac{3}{2}$ is above $\frac{1}{x}$ and find the points where the two curve intersect 2,-1/2 and zero since at zero 1/x change see this pic take the intervals where x-3/2 is above 1/x

Attachment 13156
• October 3rd 2009, 07:43 AM
unstopabl3
Thanks for your reply but I did not understand the first part at all.

The second part I understood somewhat. I didn't get what why you included 0 as an intersection point for the line and the curve and I didn't understand how we are supposed to shade/select the required region to fulfill the inequality.
• October 3rd 2009, 09:17 AM
Amer
Quote:

Originally Posted by unstopabl3
Thanks for your reply but I did not understand the first part at all.

The second part I understood somewhat. I didn't get what why you included 0 as an intersection point for the line and the curve and I didn't understand how we are supposed to shade/select the required region to fulfill the inequality.

I did not say it is a point of intersection but as you can see at zero 1/x have a disconnect point , 1/x is not continues at 0 , 1/x at 0 change it is graph as you can see .

in general you want the interval where the line is above the curve .

as in graph at the interval (-infinity , -1/2 ) the curve is above the line this interval dose not fulfill the inequality

in the interval (-1/2 , 0 ) the line is above the curve and this fulfill the inequality .

the interval (0 ,2) the curve is above the line this dose not fulfill

(2, infinity ) line is above this fulfill .

what I want to say in the first question , to solve this k(k-8)>0 we should study the sign of k(k-8) where this function is positive and where is negative after we determine this we take the interval where the function is positive , to determine this first we should find the point where is function equal zero and where it is dose not exist the zeros of the denominator , since usually the function change it is sign at these points , ok in our function we have 0,8 zero of it , we will take a number in each interval here we have three intervals (-infinity , 0) , (0,8) and (8, infinity ) ok take -1 in first interval and substitute it in the function is the sign is positive that mean the function is positive in all the first interval , do the same thing to the other intervals .

I wish it is clear now .
• October 3rd 2009, 10:20 AM
unstopabl3
Thanks much appreciated ;)