Solution of Quadratic equation

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 = $\frac{5\pm \,\sqrt(13)}{2}$


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

Condition for roots of quadratic equation.


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