# Finding for k in quadratic equation.

1. Aug 23, 2011

### Sumedh

1. The problem statement, all variables and given/known data

Find the least integral value of k for which the quadratic polynomial
(k-2)x2 + 8x + k+4 > 0 where x is real.

3. The attempt at a solution

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. Aug 23, 2011

### SteamKing

Staff Emeritus
Show us what you have done.

3. Aug 23, 2011

### 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

after that what should I do

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

4. Aug 23, 2011

### eumyang

No, it's not... at least the way I'm interpreting the question.

You are looking for the smallest integer k such that, if you plug in ANY REAL NUMBER for x, the quadratic becomes positive. That means that the graph of the parabola would be entirely above the x-axis. What does that say about the discriminant?

5. Aug 23, 2011

### SteamKing

Staff Emeritus
You might want to check your algebra on trying to solve for k in the inequality. What happened to the constant 64?

6. Aug 23, 2011

### eumyang

The algebra is actually correct. The OP multiplied the binomials, combined like terms, and then divided both sides by -4. The inequality symbol used in the beginning is wrong, however.

7. Aug 23, 2011

### Sumedh

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)
????

Last edited: Aug 23, 2011
8. Aug 23, 2011

### Sumedh

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