Solve the equation : (x^2+14x+24)(x^2+11x+24)=4x^2 ?

1 Answer
Nov 3, 2017

The Solution Set is {-1.82, -13.18, -6, -4}.

Explanation:

The given eqn. is, (x^2+14x+24)(x^2+11x+24)=4x^2.

Let, x^2+24=y. Then,

(y+14x)(y+11x)=4x^2.

:. y^2+(14x+11x)y+(14x)(15x)-4x^2=0, i.e.,

y^2+25xy+154x^2-4x^2=0, or,

y^2+25xy+150x^2=0.

:. ul(y^2+15xy)+ul(10xy+150x^2)=0.

:. y(y+15x)+10x(y+15x).

:. (y+15x)(y+10x)=0.

Since, y=x^2+24, we have,

:. (x^2+15x+24)(x^2+10x+24)=0.

If, x^2+15x+24=0, then, using the quadratic formula,

x=[-15+-sqrt{15^2-4*1*24}]/(2*1)=[-15+-sqrt(225-96)]/2.

:. x=(-15+-sqrt129)/2~~(-15+-11.36)/2, or,

x~~-1.82, or, x~~-13.18.

If, x^2+10x+24=0," then, "(x+6)(x+4)=0.

:. x=-6, or, x=-4.

Hence, the Solution Set is {-1.82, -13.18, -6, -4}.