1. ## Finding for k in quadratic equation.

Find the least integral value of k for which the quadratic polynomial
(k - 2)x^2 + 8x + k+4 > 0 where x is real.

i am trying to solve the discriminant by equating it to>0
D>0
but i don't think it is correct.
Please provide hints for this solutions.

2. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
Find the least integral value of k for which the quadratic polynomial
(k - 2)x^2 + 8x + k+4 > 0 where x is real.

i am trying to solve the discriminant by equating it to>0
D>0
but i don't think it is correct.
Please provide hints for this solutions.
I would complete the square...

3. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
Find the least integral value of k for which the quadratic polynomial
(k - 2)x^2 + 8x + k+4 > 0 where x is real.
If $\displaystyle \Delta$ is the discriminant make $\displaystyle \Delta <0$ so there are no real roots. Then look at the graph to see where it is positive.

4. ## Re: Finding for k in quadratic equation.

i did in this way

d=discriminant

D=b2-4ac
=64-4(k-2)(k+4)

as x is real
D>0

.'. 64-4(k-2)(k+4)>0

on solving
i got
(k-4)(k+6)<0

after that what should I do

should i find the values of k from this inequality???

5. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
D=b2-4ac
=64-4(k-2)(k+4) as x is real
D>0
.'. 64-4(k-2)(k+4)>0
on solving I got (k-4)(k+6)<0
So $\displaystyle k<-6\text{ or }k>4$.
I suggest that in each of those cases you make a graph of an example.
You want the graph to be above the x-axis.

6. ## Re: Finding for k in quadratic equation.

but the final answer is k=5

7. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
but the final answer is k=5
The question clearly asks for the least integer that works.
What is the least integer greater than 4?

8. ## Re: Finding for k in quadratic equation.

but why it cannot be -7 or - infinity or some integer less than -6

9. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
but why it cannot be -7 or - infinity or some integer less than -6
GRAPH the function for $\displaystyle k=-7.$ Is it positive everywhere?

10. ## Re: Finding for k in quadratic equation.

i think that on solving (k-4)(k+6)<0
we get -6<k<4

???

11. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
i think that on solving (k-4)(k+6)<0
we get -6<k<4
You have made some algebra errors somewhere.

Solve $\displaystyle 8^2-4(k-2)(k+4)<0$ and you get $\displaystyle k<-6\text{ or }4<k~.$
Typo is fixed.

12. ## Re: Finding for k in quadratic equation.

no it is not ...(k+1)
it is ...(k+4)

i think that i have to first find the zeros of the following

(k-4)(k+6)<0

that is k=-6 and k=4

then try putting any value less than -6 then value between -6 and 4
and at last value greater than 4
thus we will get the solution as -6<k<4 ???

13. ## Re: Finding for k in quadratic equation.

Originally Posted by sumedh
no it is not ...(k+1)
it is ...(k+4)
That was a TYPO.
It is now fixed. See reply #11

14. ## Re: Finding for k in quadratic equation.

the smallest value of the function:$\displaystyle f(x)=ax^2+bx+c=0$

here$\displaystyle f(x)=(k-2)x^2+8x+k+4>0$
is

$\displaystyle \frac {-D}{4a}$

to get this value positive D should be negative-------------(I)
and solving according this we get
64-4(k-2)(k+4) < 0
as this may result in correct answer

but as x is real .'. D should be positive or zero--------------(II)
$\displaystyle (I)\neq(II)$

I am confused with (I) (II)
????

15. ## Re: Finding for k in quadratic equation.

by using this

$\displaystyle \frac{-D}{4a}$

before i was running on the wrong concept