Question
How many real roots are possible for the below equation?
(x+1)(x+2)(x+3)(x+4)=24
(x+1)(x+2)(x+3)(x+4)=24
Answer: Option B
:
B
[(x+1)(x+4)][(x+2)(x+3)]=24
(x2+5x+4)(x2+5x+6)=24
Letx2+5x=k
(k+4)(k+6)=24
k2+10k=0
k(k+10)=0
k=0,k=−10
x2+5x=0,x2+5x=−10
x(x+5)=0,x2+5x+10=0
x=0,x=−5;D=25−40=−15 No real roots.
So, roots are 0, -5.
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:
B
[(x+1)(x+4)][(x+2)(x+3)]=24
(x2+5x+4)(x2+5x+6)=24
Letx2+5x=k
(k+4)(k+6)=24
k2+10k=0
k(k+10)=0
k=0,k=−10
x2+5x=0,x2+5x=−10
x(x+5)=0,x2+5x+10=0
x=0,x=−5;D=25−40=−15 No real roots.
So, roots are 0, -5.
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