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Question

The number of real roots of the equation (x+1)(x+2)(x+3)(x+4)=120 is

A
0
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B
2
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C
3
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D
4
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Solution

The correct option is B 2
Given equation
(x+1)(x+2)(x+3)(x+4)=120
(x+1)(x+4)(x+2)(x+3)=120(x2+5x+4)(x2+5x+6)=120

Assuming x2+5x=t
(t+4)(t+6)=120t2+10t96=0(t+16)(t6)=0t=6 or t=16
Now,
x2+5x=16x2+5x+16=0D=2564<0
Non real roots.

x2+5x6=0(x+6)(x1)=0x=6,1

Hence, the number of real roots are 2.

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