relationships between roots and coefficients (1 Viewer)

19KANguy · Feb 24, 2017

I have no idea how to do this question:

If a and b are roots of the equation x^2 + 8x - 5 = 0, find the equation whose roots are a/b and b/a without finding a and b.

I have to form a new equation to do this (ax^2 + bx + c = 0)

InteGrand · Feb 24, 2017

19KANguy said:
I have no idea how to do this question:

If a and b are roots of the equation x^2 + 8x - 5 = 0, find the equation whose roots are a/b and b/a without finding a and b.

I have to form a new equation to do this (ax^2 + bx + c = 0)

$\noindent Hint: find the sum $S$ and product $P$ of the roots of the desired quadratic equation, then the desired equation will be $x^2 - Sx + P = 0$ (or any non-zero multiple of this).$

integral95 · Feb 24, 2017

Yes you don't need to find a and b individually, but you'll need to find the values of

$a^2 + b^2$

$ab$

1729 · Feb 24, 2017

19KANguy said:
I have no idea how to do this question:

If a and b are roots of the equation x^2 + 8x - 5 = 0, find the equation whose roots are a/b and b/a without finding a and b.

I have to form a new equation to do this (ax^2 + bx + c = 0)

$$\noindent Now $a + b = -8$ and $ab =-5$. The roots $\left(\frac{a}{b}\right)$ and $\left(\frac{b}{a}\right)$ have a sum $\left(\frac{a}{b}\right) + \left(\frac{b}{a}\right) = \frac{a^2+b^2}{ab} = \frac{(a+b)^2-2ab}{ab} = \frac{(-8)^2-2(-5)}{-5} = \frac{-74}{5}$. The roots have a product $\left( \frac{a}{b}\right) \left(\frac{b}{a}\right) = 1$. The required quadratic equation in $x$ is of the form $x^2 - Sx + P = 0$, where $S = -\frac{74}{5}$ and $P = 1$. Thus, $x^2 + \frac{74}{5}x + 1 = 0 \Rightarrow 5x^2 + 74x + 5 = 0.$

19KANguy · Feb 24, 2017

19KANguy · Feb 24, 2017

Another one i am unsure of:

If a and b are roots of the equation x^2 + mx + n = 0, find the roots of nx^2 + (2n - m^2)x + n = 0 in terms of a and b.

1729 · Feb 25, 2017

19KANguy said:
Another one i am unsure of:

If a and b are roots of the equation x^2 + mx + n = 0, find the roots of nx^2 + (2n - m^2)x + n = 0 in terms of a and b.

$$\noindent Now $a + b = -m$ and $ab = n$. Using the quadratic formula, \\ \begin{align*} x &= \frac{m^2 - 2n \pm \sqrt{(2n-m^2)^2-4n^2}}{2n} \\ &= \frac{(a+b)^2 - 2ab \pm \sqrt{(2n-m^2-2n)(2n-m^2+2n)}}{2ab} \\ &= \frac{a^2+b^2 \pm \sqrt{-m^2(4n-m^2)}}{2ab} \\ &= \frac{a^2 + b^2 \pm m\sqrt{m^2-4n}}{2ab} \\ &= \frac{a^2+b^2 \pm (a+b)\sqrt{(a+b)^2 - 4ab}}{2ab} \\ &= \frac{a^2+b^2 \pm (a+b)\sqrt{(a-b)^2}}{2ab} \\ &= \frac{a^2 + b^2 \pm (a+b)(a-b)}{2ab} \\ &= \frac{a^2 + b^2 \pm (a^2 - b^2)}{2ab} \\ &= \frac{a}{b} \textrm{, } \frac{b}{a} \end{align*}$

19KANguy · Feb 25, 2017

thx again

zhu0902 · Feb 25, 2017

This guide is pretty extensive Good work

19KANguy · Feb 25, 2017

Solve the equation 4x^3 + 32x^2 + 79x + 60 = 0 given that one root is equal to the sum of the other two

Is there another way to do it without using factor R on Factor S?

1729 · Feb 25, 2017

19KANguy said:
Solve the equation 4x^3 + 32x^2 + 79x + 60 = 0 given that one root is equal to the sum of the other two

Is there another way to do it without using factor R on Factor S?

$\noindent Let the roots be $\alpha , \beta , \gamma \Rightarrow \alpha = \beta + \gamma.$ Now $\alpha + \beta + \gamma = 2\alpha = -8 \Rightarrow \alpha = -4.$ Thus, $(x + 4)$ is a factor of $4x^3 + 32x^2 + 79x + 60 = 0$. The other two roots can be found through long division and solving a quadratic equation in $x.$

relationships between roots and coefficients (1 Viewer)

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