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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

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 ]

Condition for roots of quadratic equation.

Polynomials