# inequality question

• Jun 25th 2013, 02:02 PM
pumbaa213
inequality question
Find the range of values of k if kx^2+8x>6-k for all real values of x

What does for all the real values of x mean?

The answer is k>8 and I can't seem to get it
this is my working by the way:

Kx^2 +8x+k-6 >0
a=k b=8 c=k-6
sub into b^2-4ac
4k^2-24k-64<0
k^2-6k-16<0
(k-8)(k+2)<0

thank you!
• Jun 25th 2013, 02:57 PM
Plato
Re: inequality question
Quote:

Originally Posted by pumbaa213
Find the range of values of k if kx^2+8x>6-k for all real values of x
The answer is k>8 and I can't seem to get it
this is my working by the way:
Kx^2 +8x+k-6 >0
a=k b=8 c=k-6
sub into b^2-4ac
4k^2-24k-64<0
k^2-6k-16<0
(k-8)(k+2)<0

It should be clear to you the inequaity is not true if $\displaystyle k<0$. Example: $\displaystyle k=-1~\&~x=10$
The parabola opens down if $\displaystyle k<0$.

On the other hand, we do need $\displaystyle b^2-4ac>0$. WHY?
• Jun 25th 2013, 08:29 PM
pumbaa213
Re: inequality question
I am sorry I still haven't managed to grasp it yet. Does real values of x means that the parabola has to be a positive and u shaped one?
And I'm not sure why we need b^2-4ac >0
• Jun 25th 2013, 09:33 PM
Prove It
Re: inequality question
Quote:

Originally Posted by pumbaa213
Find the range of values of k if kx^2+8x>6-k for all real values of x

What does for all the real values of x mean?

The answer is k>8 and I can't seem to get it
this is my working by the way:

Kx^2 +8x+k-6 >0
a=k b=8 c=k-6
sub into b^2-4ac
4k^2-24k-64<0
k^2-6k-16<0
(k-8)(k+2)<0

thank you!

\displaystyle \displaystyle \begin{align*} k\,x^2 + 8x &> 6 - k \\ k\,x^2 + 8x + k - 6 &> 0 \\ k \left( x^2 + \frac{8}{k}\,x + 1 -\frac{6}{k} \right) &> 0 \\ k \left[ x^2 + \frac{8}{k} \, x + \left( \frac{4}{k} \right) ^2 - \left( \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} \right] &> 0 \\ k \left[ \left( x + \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} - \frac{16}{k^2} \right] &> 0 \end{align*}

Notice you have the product of two things being positive. This can only happen if both things are positive or both things are negative.

Case 1: \displaystyle \displaystyle \begin{align*} k > 0 \end{align*} and \displaystyle \displaystyle \begin{align*} \left( x + \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} - \frac{16}{k^2} > 0 \end{align*}. Working on the second inequality gives

\displaystyle \displaystyle \begin{align*} \left( x + \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} - \frac{16}{k^2} &> 0 \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{16}{k^2} + \frac{6}{k} - 1 \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{16 + 6k - k^2}{k^2} \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{-\left( k^2 - 6k - 16 \right) }{k^2} \\ \left( x + \frac{4}{k }\right) ^2 &> \frac{- \left[ k^2 - 6k + (-3) ^2 - (-3 ) ^2 - 16 \right] }{k^2} \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{- \left[ \left( k - 3 \right) ^2 - 25 \right] }{k^2} \\ \left( x + \frac{4}{k } \right) ^2 &> \frac{25 - \left( k - 3 \right) ^2 }{k^2} \\ \left| x + \frac{4}{k} \right| &> \frac{\sqrt{25 - \left( k - 3 \right) ^2 }}{k} \end{align*}

Now clearly this can only have solutions if \displaystyle \displaystyle \begin{align*} 25 - \left( k - 3 \right) ^2 > 0 \end{align*}, so

\displaystyle \displaystyle \begin{align*} 25 - \left( k - 3 \right) ^2 &> 0 \\ 25 &> \left( k - 3 \right) ^2 \\ \left( k - 3 \right) ^2 &< 25 \\ | k - 3 | &< 5 \\ -5 < k - 3 &< 5 \\ -2 < k &< 8 \end{align*}

So putting the inequalities together, we find that \displaystyle \displaystyle \begin{align*} 0 < k < 8 \end{align*}.

Now try Case 2, the situation where both quantities have to be negative...
• Jun 26th 2013, 07:34 AM
Plato
Re: inequality question
Quote:

Originally Posted by pumbaa213
I am sorry I still haven't managed to grasp it yet. Does real values of x means that the parabola has to be a positive and u shaped one? And I'm not sure why we need b^2-4ac >0

For the quadratic equation $\displaystyle ax^2+bx+c=0$ we do call $\displaystyle \Delta=b^2-4ac$ is the discriminant.
The name comes from the fact it discriminates between real and complex roots.

If $\displaystyle \Delta <0$ the quadratic has no real roots. That means the entire graph is either above the x-axis or below the x-axis.

Let's review the OP
Quote:

Originally Posted by pumbaa213
Find the range of values of k if kx^2+8x>6-k for all real values of x
kx^2 +8x+k-6 >0

Here $\displaystyle \Delta=64-4(k)(k-6)$. If $\displaystyle k<0$ we see $\displaystyle 6-k>0$ so there are no real roots.

So this leaves one possible solution set: $\displaystyle k>8$
• Jul 12th 2013, 07:36 PM
pumbaa213
Re: inequality question
Thanks everyone for the help!