Roots of quadratic equations

[Lemonmelon]

I'm having trouble understanding "roots", could someone help me with this problem?

In the quadratic equation (k-1)x^2 - 5x + 3k + 4 = 0 , the roots are reciporcals of each other. Find the value of k.

usually I would apply the formula x^2 -(A + B)x + AB = 0 (where A = alpha and B = beta), however I'm not sure what to do given the (k - 1) before the x in the question..

 

[Lemonmelon]

roots of quadratic equations

In the quadratic equation (k-1)x^2 - 5x + 3k + 4 = 0, the roots are recriprocals of each other. Find the value of k.

usually I would apply the formula x^2 - (A + B)x + c = 0 to find k, however I'm not sure how to do this given the (k-1) before the x in the question. I tried to deduce k from A + B = -b/a and AB = c/a but this didn't work either... Help would be greatly appreciated .

*edit* the reason this question is posted twice is i got confused and thought the the previous thread didn't post, sorry. If anyone knows how to delete threads please notify me.

 

[ksdhart]

Hmm... unless I've done the math wrong (which is entirely possible), this equation is unsolvable. Any quadratic takes the form of \(\displaystyle ax^2+bx+c=0\). So, in this case, a is (k-1), b is -5, and c is (3k+4). Further, we're told that the two roots of the quadratic are reciprocals of one another, so the factoring will take this form:

\(\displaystyle \left(\alpha x\pm n\right)\left(\beta x\pm \frac{1}{n}\right)=0\)

Since the two roots are reciprocals, they will always have the same sign. That leaves me with the conclusion that \(\displaystyle n\cdot \frac{1}{n}=c=3k+4\) so 3k + 4 = 1 and thus k must be -1. Now, if we plug that value of k back into the original quadratic, we get:

\(\displaystyle \left(-1-1\right)x^2-5x+3\left(-1\right)+4=0\)

\(\displaystyle -2x^2-5x+1=0\)

Factoring that, however, does not yield the expected result of the two roots being reciprocals of one another. So, like I said - either I did the math wrong somewhere or the the problem is unsolvable. Maybe double check that you copied it from your book/worksheet/whatever correctly?

 

[stapel]

In the quadratic equation (k-1)x^2 - 5x + 3k + 4 = 0, the roots are recriprocals of each other. Find the value of k.

usually I would apply the formula x^2 - (A + B)x + c = 0 to find k....

I'm sorry, but I don't understand what you mean by this last line...?

Instead, try using the Quadratic Formula and the provided information. Plugging into the Formula, we get:

. . . . .\(\displaystyle x\, =\, \dfrac{-(-5)\, \pm\, \sqrt{\strut \,(-5)^2\, -\, 4(k\, -\,1)(3k\, +\, 4)\,}}{2(k\, -\, 1)}\)

. . . . .\(\displaystyle x\, =\, \dfrac{5\, \pm\, \sqrt{\strut \,25\, -\, 4(3k^2\, +\, k\, -\, 4)\,}}{2k\, -\, 2}\)

. . . . .\(\displaystyle x\, =\, \dfrac{5\, \pm\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\)

We are given that these two roots are reciprocals, so:

. . . . .\(\displaystyle \left(\dfrac{5\, -\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\right)\, \left(\dfrac{2k\, -\,2}{5\, +\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}\right)\, =\, 1\)

. . . . .\(\displaystyle 5\, -\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}\, =\, 5\, +\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}\)

For these to be equal, obviously the values of the square roots must be zero. For this to be true, the innards of the radicals must be zero. This gives us:

. . . . .\(\displaystyle 12k^2\, +\, 4k\, -\, 41\, =\, 0\)

. . . . .\(\displaystyle k\, =\, \dfrac{-(4)\, \pm\, \sqrt{\strut \, (4)^2\, -\, 4(12)(-41)\,}}{2(12)}\)

. . . . .\(\displaystyle k\, =\, \dfrac{-4\, \pm\, \sqrt{\strut \, 16\, +\, 1968\,}}{24}\)

. . . . .\(\displaystyle k\, =\, \dfrac{-4\, \pm\, \sqrt{\strut \, 1984\,}}{24}\, =\, \dfrac{-4\, \pm\, 8\, \sqrt{\strut \, 31\,}}{24}\, =\, \dfrac{-1\, \pm\, 2\, \sqrt{\strut\, 31\,}}{6}\)

Check my work, and check that these values for k actually do solve the equation.

 

[Ishuda]

In the quadratic equation (k-1)x^2 - 5x + 3k + 4 = 0, the roots are recriprocals of each other. Find the value of k.

usually I would apply the formula x^2 - (A + B)x + c = 0 to find k, however I'm not sure how to do this given the (k-1) before the x in the question. I tried to deduce k from A + B = -b/a and AB = c/a but this didn't work either... Help would be greatly appreciated .

*edit* the reason this question is posted twice is i got confused and thought the the previous thread didn't post, sorry. If anyone knows how to delete threads please notify me.

Write the equation as
a x² + b x + c = 0
with solutions
\(\displaystyle x_0\, =\, \frac{-b\, +\, \sqrt{b^2\, -\, 4\, a\, c}}{2\, a}\)
and
\(\displaystyle x_1\, =\, \frac{-b\, -\, \sqrt{b^2\, -\, 4\, a\, c}}{2\, a}\)
Since x₀ and x₁ are reciprocals we have
\(\displaystyle \frac{-b\, +\, \sqrt{b^2\, -\, 4\, a\, c}}{2\, a}\, =\, \frac{2\, a}{-b\, -\, \sqrt{b^2\, -\, 4\, a\, c}}\)
or
\(\displaystyle [-b\, +\, \sqrt{b^2\, -\, 4\, a\, c}]\, [-b\, -\, \sqrt{b^2\, -\, 4\, a\, c}]\, =\, 4\, a^2\)
or
\(\displaystyle b^2\, -\, [b^2\, -\, 4\, a\, c] =\, 4\, a^2\)
or
...
or, since a = k - 1 and c = 3 k + 4, k = ...

 

[Lemonmelon]

I'm sorry, but I don't understand what you mean by this last line...?

Instead, try using the Quadratic Formula and the provided information. Plugging into the Formula, we get:

. . . . .\(\displaystyle x\, =\, \dfrac{-(-5)\, \pm\, \sqrt{\strut \,(-5)^2\, -\, 4(k\, -\,1)(3k\, +\, 4)\,}}{2(k\, -\, 1)}\)

. . . . .\(\displaystyle x\, =\, \dfrac{5\, \pm\, \sqrt{\strut \,25\, -\, 4(3k^2\, +\, k\, -\, 4)\,}}{2k\, -\, 2}\)

. . . . .\(\displaystyle x\, =\, \dfrac{5\, \pm\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\)

We are given that these two roots are reciprocals, so:

. . . . .\(\displaystyle \left(\dfrac{5\, -\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\right)\, \left(\dfrac{2k\, -\,2}{5\, +\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}\right)\, =\, 1\)

. . . . .\(\displaystyle 5\, -\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}\, =\, 5\, +\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}\)

For these to be equal, obviously the values of the square roots must be zero. For this to be true, the innards of the radicals must be zero. This gives us:

. . . . .\(\displaystyle 12k^2\, +\, 4k\, -\, 41\, =\, 0\)

. . . . .\(\displaystyle k\, =\, \dfrac{-(4)\, \pm\, \sqrt{\strut \, (4)^2\, -\, 4(12)(-41)\,}}{2(12)}\)

. . . . .\(\displaystyle k\, =\, \dfrac{-4\, \pm\, \sqrt{\strut \, 16\, +\, 1968\,}}{24}\)

. . . . .\(\displaystyle k\, =\, \dfrac{-4\, \pm\, \sqrt{\strut \, 1984\,}}{24}\, =\, \dfrac{-4\, \pm\, 8\, \sqrt{\strut \, 31\,}}{24}\, =\, \dfrac{-1\, \pm\, 2\, \sqrt{\strut\, 31\,}}{6}\)

Check my work, and check that these values for k actually do solve the equation.

I think the curriculum where I am (Australia) is different to that of where you guys are because we haven't used the quadratic equation in relation to roots.. However, i hadn't realised that roots multiplied equals 1... thus I can apply AB = c/a to find k. Thanks for the help

 

[Cartesius24]

little bitvof solution

1. First our equation has to have two roots so \(\displaystyle \Delta >0 \)it can't be forgotten.
2. Also \(\displaystyle x_{1} \cdot x_{2} = 1 \) as they are reciprocals. So we have from 2) equation wich solution has to be checked for condition 1) Any further infos?

 

[stapel]

1. First our equation has to have two roots so \(\displaystyle \Delta >0 \)it can't be forgotten.
2. Also \(\displaystyle x_{1} \cdot x_{2} = 1 \) as they are reciprocals. So we have from 2) equation wich solution has to be checked for condition 1) Any further infos?

What? Is this a new question of some sort? If not, how does this relate to the question at hand?

 

[stapel]

I'm sorry, but I don't understand what you mean by this last line...?

Instead, try using the Quadratic Formula and the provided information. Plugging into the Formula, we get:

. . . . .\(\displaystyle x\, =\, \dfrac{-(-5)\, \pm\, \sqrt{\strut \,(-5)^2\, -\, 4(k\, -\,1)(3k\, +\, 4)\,}}{2(k\, -\, 1)}\)

. . . . .\(\displaystyle x\, =\, \dfrac{5\, \pm\, \sqrt{\strut \,25\, -\, 4(3k^2\, +\, k\, -\, 4)\,}}{2k\, -\, 2}\)

. . . . .\(\displaystyle x\, =\, \dfrac{5\, \pm\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\)

We are given that these two roots are reciprocals, so:

. . . . .\(\displaystyle \left(\dfrac{5\, -\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\right)\, \left(\dfrac{2k\, -\,2}{5\, +\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}\right)\, =\, 1\). . .<==. .Wrong! Wrong! Wrong!

Thank you, Ishuda, for the kind and gentle correction via private message. You're quite right: being reciprocals already, neither of the roots needs to be flipped in order to create a product of "1".

So, starting over from the error, we get:

. . . . .\(\displaystyle \left(\dfrac{5\, -\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\right)\,\left(\dfrac{5\, +\, \sqrt{\strut \, 41\, -\, 4k\, -\, 12k^2\,}}{2k\, -\, 2}\right)\,=\, 1\)

The two terms which contain radicals cancel out, so the product gives us:

. . . . .\(\displaystyle \dfrac{25\, -\, \left(41\, -\, 4k\, -\, 12k^2\right)}{4k^2\, -\, 8k\, +\, 4}\, =\, 1\)

. . . . .\(\displaystyle 12k^2\, +\, 4k\, -\, 16\, =\, 4k^2\, -\, 8k\, +\, 4\)

. . . . .\(\displaystyle 8k^2\, +\, 12k\, -\, 20\, =\, 0\)

. . . . .\(\displaystyle 2k^2\, +\, 3k\, -\, 5\, =\, 0\)

. . . . .\(\displaystyle (2k\, +\, 5)\,(k\, -\, 1)\, =\, 0\)

Etc, etc, and so forth.