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

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Answer 1 · 2017-11-03T18:52:30

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

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}.#