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How many real solutions of the equation 6x^(2)-77[x]+147=0, where [x] is the integral part of x? |
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Answer» `implies(6x^(2)+147)/77=[x]` `(0.078)x^(2)=[x]-1.9` `:'(0.078)x^(2)gt0=x^(2)gt0` `:.[x]-1.9gt0` or `[x]gt1.9` `:.[x]=2,3,4,5`………… If `[x]=2` i.e. `2lexlt3` Then `x^(2)=(2-1.9)/0.078=1.28` `:.x=1.13` [fail] If `[x]=3,` i.e. `3lexlt4` Then `x^(2)=(3-1.9)/0.078=14.1` `:.x=3.75` [true] If `[x]=4`, i.e. `4lexlt5` Then `x^(2)=(4-1.9)/(0.078)=26.9` `:.x=5.18` ..[fail] If `[x]=5,` i.e. `5lexlt6` Then `x^(2)=(5-1.9)/0.078=39.7` `:.x=6.3` [fail] If `[x]=6,` i.e. `6lexlt7` Then `x^(2)=(6-=1+N3269)/0.078=4.1/0.078=52.56` `:.x==.25`[fail] If `[x]=7`, i.e. /7lexlt8` Then `x^(2)=(7-1.9)/0.078=5.1/0.078=65.38` `:.x=8.08` [fail] If `[x]=8` i.e. `8lexlt9` Then `x^(2)=(8-1.9)/0.078=6.1/0.078=78.2` `:.x=8.8` [true] If `[x]=9,` i.e. `9lexlt10` Then `x^(2)=(9-1.9)/0.078=7.1/0.078=91.03` `:.x=9.5` [true] If `[x]=10`, i.e `10lexlt11` Then `x^(2)=(10-1.9)/0.078=8.1/0.078=103.8` `:.x=10.2` If `[x]=11`, i.e. `11lexlt12` Then `x^(2)=(11-1.9)/(0.078` `=9.1/0.078=116.7` `:.x=10.8` [fail] Other values of fail. hence number of SOLUTIONS is four. |
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