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Baby-Nath24
Solve the inequality: x^2 - 5x + 6 ≥ 0. A. x ≤ 2 or x ≥ 3 B. 2 ≤ x ≤ 3 C. x ≤ or x ≥ 6 D. x ≤ -6 or x ≥ -1
(x-3)(x-2)= x^2-5x+6, do you need more help?
\[x ^{2}-5x+6>0,x ^{2}-5x>-6\] adding both sides \[\left( \frac{ 5 }{ 2 } \right)^{2} i.e.,\frac{ 25 }{ 4 }\] \[x ^{2}-5x+\frac{ 25 }{4 }>-6+\frac{ 25 }{4 }\] \[\left( x-\frac{ 5 }{ 2 } \right)^{2}>\frac{ 1 }{ 4 },\left| x-\frac{ 5 }{2 } \right|>\frac{ 1 }{ 2 }\] solve it. \[(\left| x-a \right|>b means x-a <-b and x-a >b)\]
(x-3)(x-2)>=0 (its equal or greater than, but didn't know how to do the sign) x-3>=0 and x-2>=0 just solve for x and see in which one it is right
\[correction everywhere write \ge inplace of >\]