H S Hall and S R Knight Solutions for Chapter: Theory of Equations, Exercise 5: EXAMPLES. XXXV. e.

Show that the equation $x^4+p{x}^3+q{x}^2+rx+s=0$ may be solved as a quadratic if $r^2=p^{2}s$.

Solve the equation $x^6-18x^4+16x^3+28x^2-32x+8=0$ one of whose roots is $\sqrt{6}-2$.

If $\alpha , \beta , \gamma , \delta$ are the roots of the equation, $x^4+q{x}^2+rx+s=0$. Find the equation whose roots are $\beta +\gamma +\delta +(\beta \gamma \delta {)}^{-1}$ and $c$.

In the equation $x^4-p{x}^3+q{x}^2-rx+s=0$, prove that if the sum of two of the roots is equal to the sum of the other two, then $p^3-4pq+8r=0$ and that if the product of two of the roots is equal to the product of the other, then $r^2=p^{2}s$.

The equation $x^5-209x+56=0$ has two roots whose product is unity, determine them.

Find the two roots of $x^5-409x+285=0$ whose sum is $5$.

If $a, b, c, \dots \dots , k$ are the $n$ roots of $x^n+p_1{x}^{n-1}+p_2{x}^{n-2}+\dots \dots +p_{n-1}x+p_n=0$. Then, show that $\left(1+a^2\right)\left(1+b^2\right)\dots \left(1+k^2\right)={\left(1-p_2+p_4-\dots \right)}^2+{\left(p_1-p_3+p_5-\dots \right)}^2$.

The sum of two roots of the equation $x^4-8x^3+21x^2-20x+5=0$ is $4.$ Solve the equation from a knowledge of this fact.