The solution of by-quadratic equation of the form
(x+a)(x+b)(x+c)(x+d)=e such that a+b=c+d=p
The above by-quadratic can be easily transformed into the quadratic equation.
Lets look into the solution of by-quadratic equation of above form.
Given a+b = c+d = p
Hence (x+a)(x+b)(x+c)(x+d)=e
$(x^2\,+(a+b)x\,+ab\,)(\,x^2\,+(c+d)x\,+cd)=\,e $
Let $ \,\,\,\,x^2\,+\,px\,=\,y $
Hence equation reduces to quadratic in y as (y+ab)(y+cd)=e
Solution of Quadratic equation is let $\,y_1\,,\,y_2\,$
Then Solving the equation $\,x^2+px=y_i\, \,\, for\,\,i=1\,,2$
For example Solution of by-quadratic equation (x-1)(x-2)(x-3)(x-4)=3
$(x^2\,-\,5x\,+\,4)(x^2\,-\,5x\,+6)\,=\,3 $ $\\$ Let y=$(x^2\,-\,5x\,+4)$ $\\$ Hence Equation will get reduced to $ y^2\,+\,2y\,=\,3 $
Hence y= -3 0r 1
When y=-3 roots are imaginary and when y=1 roots are x = $\frac{5\pm \,\sqrt(13)}{2}$
ANOTHER APPLICATION IS TO FIND RANGE
Find range of f(x)=(x-1)(x-2)(x-3)(x-4) for x$\in \,$ [0 , 6]
Hence f(y) = y(y+2) where y = $x^2\,-\,5x\,+4\, $ $\\$ For x $\in \, $[0,6]
y $\in \,[\frac{-9}{4} \, ,\,10\,] $
Hence f(x) $\in \, $ [-1 , 120 ]
Note :- (1) Condition for roots of quadratic equation $ax^2+bx+c=0 \,where\,\,,a\neq 0\,\,,\, a,b,c \in R$
(2) For relation between coefficients and zeros of polynomial