Solve $(x+1)(x+3)(x+4)(x+6)=112$ ?

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Answer 1 · 2017-06-11T08:55:09

$x=-7/2+-isqrt31/2$ or $x=-7/2+-sqrt57/2$

Let us group LHS as

$(x+1)(x+6)(x+3)(x+4)=112$

$=>(x^2+7x+6)(x^2+7x+12)=112$

Now let $u=x^2+7x$ and then above equation becomes

$(u+6)(u+12)=112$

or $u^2+18u+72=112$

or $u^2+18u-40=0$

or $(u+20)(u-2)=0$ i.e. $u=2$ or $-20$

As such either $x^2+7x+20=0$ i.e. $x=(-7+-sqrt(7^2-80))/2$ i.e. $x=-7/2+-isqrt31/2$

or $x^2+7x-2=0$ i.e. $x=(-7+-sqrt(7^2+8))/2$ i.e. $x=-7/2+-sqrt57/2$

Solve (x+1)(x+3)(x+4)(x+6)=112(x+1)(x+3)(x+4)(x+6)=112?