Tricky Quadratic.

[Chrispy]

Here's another one

x²+5x-1-k(x²+1)=0

Find values between which k must lie for x to have real roots. Then find values of k for which one root is twice the other.
Not sure if there are holes in my basic algebra or if I'm just thinking about it all wrong.

 

[Chrispy]

Tricky quadratic

I did this:

(1-k)x² +5x-(1+k) = 0

Got to here:


x= -5±√2̅̅5̅-̅4̅(̅k̅²̅-̅1̅)̅
̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅2̅(̅1̅-̅k̅)̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ now stuck.

 

[Deleted member 4993]

I did this:

(1-k)x² +5x-(1+k) = 0

Got to here:


x= -5±√2̅̅5̅-̅4̅(̅k̅²̅-̅1̅)̅
̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅2̅(̅1̅-̅k̅)̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ now stuck.

Now for x to be real, the quantity under radical must be positive. So,

25 - 4k² + 4 ≥ 0

4k² ≤ 29

Now calculate "k"

 

[lookagain]

Now for x to be real, the quantity under radical must be positive. \(\displaystyle \ \ \)non-negative

.

 

[Chrispy]

Thanks for clarifying guys.

 

[Deleted member 4993]

.

You are correct - in general. However, for this problem the roots needed to be unique (non-repeating - one twice the other) and real - thus the discriminant needed to be positive.

 

[Chrispy]

Now for x to be real, the quantity under radical must be positive. So,

25 - 4k² + 4 ≥ 0

4k² ≤ 29

Now calculate "k"

Great thanks.

So I get this: k ≤ ± √29/2

but obviously k lies between those two values (inclusive). How do I resolve the contradiction without losing clarity of expression?

 

[Deleted member 4993]

Great thanks.

So I get this: k ≤ ± √29/2

but obviously k lies between those two values (inclusive). How do I resolve the contradiction without losing clarity of expression?

What is the contradiction you are observing here?

The output is |k| ≤ √(29)/2

Now apply the other given condition → One root is twice that of the other.

 

[Dale10101]

Perspective

Here's another one

x²+5x-1-k(x²+1)=0

Find values between which k must lie for x to have real roots. Then find values of k for which one root is twice the other.
Not sure if there are holes in my basic algebra or if I'm just thinking about it all wrong.

Not really anything new here, but ... from other problems I have seen the concise solution path focus' on the "discriminant" rather then the quadratic formula.

As before:

\[{{\rm{x}}^{\rm{2}}} + {\rm{5x}} - {\rm{1}} - {\rm{k}}({{\rm{x}}^{\rm{2}}} + {\rm{1}}) = 0\,\,\,\, = > \,\,\,\,(1 - k){x^2} + 5x + ( - k - 1) = 0\]

except that the standard format requires the constant at the end written as shown above to allow stating:

\[\begin{array}{l}
a = (1 - k)\\
b = 5\\
c = ( - k - 1)
\end{array}\]

D = the discriminant
\[D = {b^2} - 4ac = {5^2} - 4(1 - k)( - k - 1)\,\,\, = > \,\,\,D = 29 - 4{k^2}\]

The facts about the discriminant:

• If discriminant (D) is equal to 0 then the equation has one real solution.
• If D > 0, then the equation has two real solutions.
• If D < 0, then the equation has two imaginary solutions.

As before, the values of k providing two real roots occur when D>0:

\[D > 0\,\,\,\,\,\, = > \,\,\,\,29 - 4{k^2} > 0\,\,\,\,\, = > k < \left| {\frac{{\sqrt {29} }}{2}} \right|\,\,\,\,\, = > \,\,\,\,\,\frac{{ - \sqrt {29} }}{2} < k < \frac{{\sqrt {29} }}{2}\]

The value of k providing a single real root occurs when D = 0:
\[D = 0\,\,\,\, = > \,\,\,\,29 - 4{k^2} = 0\,\,\,\,\, = > \,\,\,\,k = \frac{{\sqrt {29} }}{2}\,\,\,\, \vee \,\,\,k = \frac{{ - \sqrt {29} }}{2}\]

A plot of k vs the value of the discriminant:

 

[Chrispy]

What is the contradiction you are observing here?

The output is |k| ≤ √(29)/2

Now apply the other given condition → One root is twice that of the other.

I may have been unclear; it was kind of a two part question. The second part was about one root being twice the other. The first part was just finding all values for k which gave x real roots.

But I think you may have answered my question by simply applying absolute value brackets.

Now, as you say ,I'll get to work on the second condition.

 

[Chrispy]

Not really anything new here, but ...

A plot of k vs the value of the discriminant:

View attachment 3699

Wow, that's very informative, thanks!

 

[Dale10101]

Nada

Great thanks.

So I get this: k ≤ ± √29/2

but obviously k lies between those two values (inclusive). How do I resolve the contradiction without losing clarity of expression?

You're Welcome, . Am a student of math myself, and many here help me too.

Regarding your question about the "contradiction":

Great thanks.

So I get this: k ≤ ± √29/2

but obviously k lies between those two values (inclusive). How do I resolve the contradiction without losing clarity of expression?

,
You did not seem too satisfied with M. Subhotosh Khan's response which was of course perfectly accurate but terse.

I think you got off on the wrong track and sort of abstracted k ≤ ± √29/2 from the quadratic formula..

I think what you wanted to do was this:
\[4{k^2} \le 29\,\,\,\, \Rightarrow \,\,\,\,{k^2} \le \frac{{29}}{4}\]
Next apply the square root operator to both sides:
\[\sqrt {{k^2}} \le \sqrt {\frac{{29}}{4}} \,\,\, = > \,\,\,\sqrt {{k^2}} \le \frac{{\sqrt {29} }}{2}\,\,\, = > \,\,\,\left| k \right| \le \frac{{\sqrt {29} }}{2}\]

The left hand side of the last inequity follows from the basic property:
\[\sqrt {{a^2}} = \left| a \right|\],

the argument of the right hand side is a positive number so it's root will be positive and not + or - , only +.

Finally, another basic property states:
\[\left| a \right| \le b\,\,\,\, \Leftrightarrow \,\,\,\,\, - b \le a \le b\]

so:
\[\left| k \right| \le \frac{{\sqrt {29} }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\frac{{ - \sqrt {29} }}{2} \le k \le \frac{{\sqrt {29} }}{2}\]

This is interesting problem and I am still studying it, there is the question of how to proceed to the second half of the problem itself but I am looking at the original equation too and trying to extract as much information about its meaning from it's properties. For example, in going from k = 0 to k >0 what sort of family of equations is introduced. Anyway, it would be interesting to see your next calculations if you have time.

 

[Deleted member 4993]

If we have a quadratic equation Ax² + Bx + C = 0 whose roots are M and N, then

M + N = -B/A

and

M * N = C/A

In this case M = 2*N

Original equation

x²+5x-1-k(x²+1)=0 → A = 1-k , B = 5 & C = -k - 1

then

3N = - 5/(1-k) & N² = -(k+1)/[2(1-k)] = (k+1)/[2(k-1)] → k > 1

N²/(3N) = (k+1)/10 → N = 3/10 * (k+1) → 9/100 * (k+1)² = (k+1)/[2(k-1)] → k² -1 = 50/9 → k = √59/3 (which is < √29/2)

Did it very quickly - may have arithmetic error - ready to go to corner.....

 

[lookagain]

---------> You are correct - in general. However, for this problem the roots needed to be unique
(non-repeating - one twice the other) and real - thus the discriminant needed to be positive.

Now for x to be real, the quantity under radical must be positive. So,

25 - 4k² + 4 ≥ 0

4k² ≤ 29

Yes, but you were inconsistent, because you mixed the word "positive" with the \(\displaystyle \ "\ge" \ \) symbol.

 

[Chrispy]

If we have a quadratic equation Ax² + Bx + C = 0 whose roots are M and N, then

M + N = -B/A

and

M * N = C/A

In this case M = 2*N

Original equation

x²+5x-1-k(x²+1)=0 → A = 1-k , B = 5 & C = -k - 1

then

3N = - 5/(1-k) & N² = -(k+1)/[2(1-k)] = (k+1)/[2(k-1)] → k > 1

N²/(3N) = (k+1)/10 → N = 3/10 * (k+1) → 9/100 * (k+1)² = (k+1)/[2(k-1)] → k² -1 = 50/9 → k = √59/3 (which is < √29/2)

Did it very quickly - may have arithmetic error - ready to go to corner.....

Which is what the book says. Wasn't doubting you earlier; I was assuming there was some convention I didn't know about. Thanks to you and Nada for great help .