Review question

# What if the roots of this equation are in geometric progression? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7111

## Solution

1. Show that, if the roots of the equation $x^3-5x^2+qx-8=0$ are in geometric progression, then $q=10$.

Since we know the roots are in geometric progression, and the leading coefficient is $1$, we have $x^3 - 5x^2 + qx - 8 = (x-c)(x-cr)(x-cr^2),$ so $x^3 - 5x^2 + qx - 8 = x^3 - c(1 + r + r^2) x^2 + c^2(r + r^2 + r^3) x - c^3 r^3,$ and this is true for all values of $x$.

Comparing coefficients of the different powers of $x$ now tells us that

1. $c^3 r^3 = 8$,
2. $c(1+r+r^2) = 5$, and finally that
3. $q = c^2r(1+r+r^2)$.

But now notice that $q = c(1+r+r^2) \times cr = 5 \times 8^{1/3} = 10,$ as desired.

1. If $\alpha$, $\beta$, $\gamma$ are the roots of the equation $x^3-x^2+4x+7=0,$ find the equation whose roots are $\beta+\gamma$, $\gamma+\alpha$, $\alpha+\beta$.

Using the same approach as above, we find that

1. $\alpha + \beta + \gamma = 1$,
2. $\alpha\beta + \beta\gamma + \gamma\alpha = 4$,
3. $\alpha\beta\gamma = -7$.

Suppose the polynomial we need is of the form $x^3 + A x^2 + B x + C = (x-\beta-\gamma)(x-\gamma - \alpha)(x-\alpha-\beta).$

Expanding gives

1. $-A = (\beta+\gamma)+(\gamma+\alpha)+(\alpha+\beta),$
2. $B = (\beta+\gamma)(\gamma + \alpha)+(\beta+\gamma)(\alpha+\beta)+(\gamma + \alpha)(\alpha+\beta),$
3. $-C = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$.

Consider first $A$. We see that $-A = (2\alpha + 2\beta + 2\gamma) = 2(\alpha+\beta+\gamma) = 2,$ so $A = -2$.

Next consider \begin{align*} B &= (\beta+\gamma)(\gamma + \alpha)+(\beta+\gamma)(\alpha+\beta)+(\gamma + \alpha)(\alpha+\beta)\\ &= \alpha^2 + \beta^2 + \gamma^2 + 3\alpha\beta+3\beta\gamma+3\gamma\alpha\\ &=\alpha^2 + \beta^2 + \gamma^2 + 2\alpha\beta+2\beta\gamma+2\gamma\alpha + (\alpha\beta+\beta\gamma+\gamma\alpha)\\ &=(\alpha+\beta+\gamma)^2+4\\ &=5. \end{align*} Before we tackle $C$, let’s note that \begin{align*} 1 \times 4 &= (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)\\ &= \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta+3\alpha\beta\gamma\\ &= \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta-21\\ &\implies \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta =25.\\ \end{align*} Now we have \begin{align*} -C &= (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)\\ &= \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta + 2 \alpha\beta\gamma\\ &= 25 -14 = 11, \end{align*}

and so $C = -11.$

Therefore the required equation is $x^3 - 2x^2 + 5x - 11=0.$