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

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Answer 1 · 2017-11-03T18:52:30

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

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

Let, $x^{2} + 24 = y .$ $x^2+24=y.$ Then,

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

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

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

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

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

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

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

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 ${-1.82, -13.18, -6, -4}.$

Solve the equation : (x^2+14x+24)(x^2+11x+24)=4x^2(x2+14x+24)(x2+11x+24)=4x2(x^2+14x+24)(x^2+11x+24)=4x^2 ?