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# Reduce equation to quadratic form

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

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

Note by Joshi Rishit
4 years ago

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You 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! · 4 years ago

no sorry brother ans is not comin any way thanx for ur help · 4 years ago

Continuing where Mursalin left:

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. · 4 years ago

Thanx very much bro · 4 years ago