Reduce equation to quadratic form

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

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

#Algebra #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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!

Mursalin Habib - 7 years, 10 months ago

no sorry brother ans is not comin any way thanx for ur help

JOSHI RISHIT - 7 years, 10 months 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.

Swaraj Yadav - 7 years, 10 months ago

@Swaraj Yadav – Thanx very much bro

JOSHI RISHIT - 7 years, 10 months 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.....

Vivek Sahu - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$