Question

(x+1)(x+2)=(x-3)(x-4)​

Answer 1

Answer:What is the solution of (x+1)(x+2)(x+3)(x+4)=120?

From

120=(x+1)(x+2)(x+3)(x+4)=((x+1)(x+4))((x+2)(x+3))

=((x2+5x+5)−1)((x2+5x+5)+1)…(1)

=(x2+5x+5)2−1,

we get x2+5x+5=±11.

Thus x2+5x−6=0 or x2+5x+16=0. The first of these give x=1 or x=−6. The second of these give x=12(−5±−39−−−−√).

It is easy to verify that x=1 and x=−6 are both solutions; in fact, 120=2⋅3⋅4⋅5=(−5)⋅(−4)⋅(−3)⋅(−2).

That the second pair of complex conjugates is also a solution can be best verified by replacing x2+5x by −16 in eqn. (1), say.

There are two real solutions, x=1 and x=−6, and two non-real solutions x=12(−5±−39−−−−√). ■

Hope it helps!!!

Step-by-step explanation:

Answer 2

Step-by-step explanation:

formula- (x+a)(x+b)=>x²+(a+b)x+ab

~

(x+1)(x+2)=(x-3)(x-4)

x²+(1+2)x+1×2=x²-(3-4)x-3×4

x²+2x+2=x²-1x-12

x²-x²+2x-1x+2-12

0+x-10

=> x-10 is the answer