Thread: [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)

Answer is 0>k>8

I solved the above to get



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)

I solved this and got:





Answer is k=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 and . Find the solution set of the inequality

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


Thanks in advance!

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)

Answer is 0>k>8

I solved the above to get



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)

I solved this and got:





Answer is k=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 and . Find the solution set of the inequality

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


Thanks in advance!

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 there is no solution right but here

for all k real number so always there is a equal roots and the root is

repeated root so there is no value of k that make no solution

iii) draw the two curve and take the region where is above 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

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.

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 .

Thanks much appreciated