maths question (1 Viewer)

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.

How do i do this without using Factor R on Factor S?

KingOfActing · Feb 25, 2017

I don't know what you mean by "Factor R on Factor S" but this is the way I would solve it:

$\text{Let the roots be }a,\,b,\text{ and }c\text{. It is given that }a = b + c$

$\begin{align*}\\\text{By the sum of roots: }\\a + b + c &= \frac{-32}{4} = -8\\2a&= -8\\a &= -4 \end{align*}$

$\\$Therefore, one of the roots is $x = -4$. Now proceeding with long division, we get that $4x^3 + 32x^2 + 79x + 60 = (x+4)(4x^2 + 16x + 15)$. Therefore the roots are:$\\x = -4,\, \frac{-16 + \sqrt{16^2 -4\times4\times15}}{2\times4} = \frac{-3}{2},\,\frac{-16 - \sqrt{16^2 -4\times4\times15}}{2\times4} = \frac{-5}{2}$

19KANguy · Feb 25, 2017

19KANguy · Feb 25, 2017

some more questions:

1. Find the cubic equation whose roots are twice those of the equation 3x^3 - 2x^2 + 1 = 0
2. Find two values of m such that the roots of the equation x^3 + 2x^2 + mx - 16 = 0 are a, b, ab. Use these values of m to find a and b

Mahan1 · Feb 26, 2017

19KANguy said:
some more questions:

1. Find the cubic equation whose roots are twice those of the equation 3x^3 - 2x^2 + 1 = 0
2. Find two values of m such that the roots of the equation x^3 + 2x^2 + mx - 16 = 0 are a, b, ab. Use these values of m to find a and b

1.

Suppose the roots of the original equation are

$\alpha , \beta, \gamma$

Using sum and products of the roots we know:

$\alpha + \beta + \gamma = \frac{2}{3}, \alpha \ \beta + \alpha \ \gamma + \beta \ \gamma = 0 , \alpha \ \beta \ \gamma = -\frac{1}{3}$

To create a polynomial which has roots that are doubled, we let

$\alpha' = 2 \alpha , \beta' = 2\beta, \gamma' = 2\gamma$

hence we get:

$\alpha' + \beta' + \gamma = 2(\alpha + \beta + \gamma) = 2(\frac{2}{3}) = \frac{4}{3}$

similarly,

$\alpha' \ \beta' + \alpha' \ \gamma '+ \beta' \ \gamma' = 4( \alpha \ \beta + \alpha \ \gamma + \beta \ \gamma )= 0$

$\alpha' \ \beta' \ \gamma' = 8(\alpha \ \beta \ \gamma )= 8(-\frac{1}{3}) = -\frac{8}{3}$

keeping the coefficient of $x^{3}$ the same we get:

$3x^{2} - 4x^{2} + 8 = 0$

2.
using product of the roots:

$ab(ab) = (ab)^{2}=16 \Rightarrow ab = \pm 4$

if ab= 4 and using the sum of the roots:

$a+b+ab = -2 \Rightarrow a+b= -6$

$ab + a(ab) + b(ab) = ab(1+a+b) = 4(-5) = -20 = m$

Similarly if ab = -4, we get:

$a+b+ab = -2 \Rightarrow a+b= 2$

$ab + a(ab) + b(ab) = ab(1+a+b) = 4(2) = 10 = m$

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maths question (1 Viewer)

19KANguy

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