# Finding for k in quadratic equation.

• Aug 23rd 2011, 07:05 AM
sumedh
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.
• Aug 23rd 2011, 07:27 AM
Prove It
Re: Finding for k in quadratic equation.
Quote:

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...
• Aug 23rd 2011, 07:31 AM
Plato
Re: Finding for k in quadratic equation.
Quote:

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 $\Delta$ is the discriminant make $\Delta <0$ so there are no real roots. Then look at the graph to see where it is positive.
• Aug 23rd 2011, 07:48 AM
sumedh
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???
• Aug 23rd 2011, 08:00 AM
Plato
Re: Finding for k in quadratic equation.
Quote:

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 $k<-6\text{ or }k>4$.
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.
• Aug 23rd 2011, 08:19 AM
sumedh
Re: Finding for k in quadratic equation.
but the final answer is k=5
• Aug 23rd 2011, 08:25 AM
Plato
Re: Finding for k in quadratic equation.
Quote:

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?
• Aug 23rd 2011, 08:28 AM
sumedh
Re: Finding for k in quadratic equation.
but why it cannot be -7 or - infinity or some integer less than -6
• Aug 23rd 2011, 08:36 AM
Plato
Re: Finding for k in quadratic equation.
Quote:

Originally Posted by sumedh
but why it cannot be -7 or - infinity or some integer less than -6

GRAPH the function for $k=-7.$ Is it positive everywhere?
• Aug 23rd 2011, 10:19 AM
sumedh
Re: Finding for k in quadratic equation.
i think that on solving (k-4)(k+6)<0
we get -6<k<4

(Thinking)???
• Aug 23rd 2011, 10:44 AM
Plato
Re: Finding for k in quadratic equation.
Quote:

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 $8^2-4(k-2)(k+4)<0$ and you get $k<-6\text{ or }4
Typo is fixed.
• Aug 23rd 2011, 10:52 AM
sumedh
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 ??? (Worried)
• Aug 23rd 2011, 10:59 AM
Plato
Re: Finding for k in quadratic equation.
Quote:

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
• Aug 23rd 2011, 06:28 PM
sumedh
Re: Finding for k in quadratic equation.
the smallest value of the function: $f(x)=ax^2+bx+c=0$

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

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

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)
$(I)\neq(II)$

I am confused with (I) (II)
????(Thinking)
• Aug 23rd 2011, 07:52 PM
sumedh
Re: Finding for k in quadratic equation.
$\frac{-D}{4a}$