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.

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...

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 $\displaystyle \Delta$ is the discriminant make $\displaystyle \Delta <0$ so there are no real roots. Then look at the graph to see where it is positive.

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???

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 $\displaystyle k<-6\text{ or }k>4$.

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.

Re: Finding for k in quadratic equation.

but the final answer is k=5

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?

Re: Finding for k in quadratic equation.

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

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 $\displaystyle k=-7.$ Is it positive everywhere?

Re: Finding for k in quadratic equation.

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

we get -6<k<4

(Thinking)???

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

Typo is fixed.

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)

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

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)

????(Thinking)

Re: Finding for k in quadratic equation.

i got the answer

by using this

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

before i was running on the wrong concept

now i got the answer

thank you very much for your valuable suggestion and valuable time for me

thank you once again