\( (X+1)(X+2)(X+3)(X+4)=120 \),

How to bring this equation to the form of \( ax^{2} +bx+c=0 \)

\( (X+1)(X+2)(X+3)(X+4)=120 \),

How to bring this equation to the form of \( ax^{2} +bx+c=0 \)

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TopNewestYou can always start by multiplying all the terms, but it will get really messy!

Here's a good way. Rearrange the terms like this:

\((X+1)(X+4)(X+2)(X+3)=120\)

And multiply two terms at a time:

\((X^2+5X+4)(X^2+5X+6)=120\)

Let \(x=X^2+5X+4\).

Now we have something like this:

\(x(x+2)=120\)

or \(x^2+2x-120=0\).

Notice that this equation is in the form \(ax^2+bx+c=0\) where \(a=1\), \(b=2\) and \(c=-120\). So we're done!

Hope this helps! – Mursalin Habib · 4 years ago

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– Joshi Rishit · 4 years ago

no sorry brother ans is not comin any way thanx for ur helpLog in to reply

after you got \(x^{2}+2x-120=0\) solve it as \(x^{2}+12x-10x-120=0\) => \(x(x+12)-10(x+12)=0\) =>\((x+12)(x-10)=0\) which gives \(x=10,-12\)

for \(x=10, X^{2}+5X+4=10 => X^{2}+5X-6=0 => X^{2}+6X-X-6=0 \) \(=> X(X+6)-(X+6)=0 => (X+6)(X-1)=0\) which gives \(X=1,-6\)

for \(x=-12, X^{2}+5X+4=-12 => X^{2}+5X+16=0\) which does not have any real roots since \(b^{2}-4ac\) i.e., \(25-64 <0\)

so the answers are 1 and -6. – Swaraj Yadav · 4 years ago

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– Joshi Rishit · 4 years ago

Thanx very much broLog in to reply

– Vivek Sahu · 4 years ago

you mean to say that arrange in such a way that the the sum of constants i.e 1,4 & 2,3 are same i.e. 5 lemme try.....Log in to reply