You can simplify the equation as follows
(x-3)(x-2)<0. M=(x+3)(x+2). We are trying to find a range for M
In order for (x-3)(x-2) to be < 0, either (x-3)>0 or (x-2)<0 . This is one case
(x-3)<0 or (x-3)>0. This is another case. Out of the two only one satisfies the condition
We have 2<x<3
Now we now M=(x+3)(x+2). Let us substitute the values of 2 and 3 respectively to find the lower and upper bound for M. We get 20<M<30. Answer is B
2)
(7+4sqrt(3))^x^2-8 + (7-4sqrt(3))^x^2-8 = 14. For this problem it is easy to substitute values from the answer choice
The first solution +-3,+-1, only partially satisfies the equation
The second solution +-3,+-sqrt(7) satisfies the equation for both values of x
Substitute x=3 or -3, it doesn't matter as we are squaring the result, we get
(7+4sqrt(3))^9-8 + (7-sqrt(3))^9-8 = 7+sqrt(4) +7-sqrt(4) = 14
Substitute x=sqrt(7), we get
(7+4sqrt(3))^7-8 + (7-4sqrt(3))^7-8
1/(7+4sqrt(3)) + 1/(7-sqrt(4)) . Cross multiply, we get 14 in the numerator and 1 in the denominator
Denominator (7+4sqrt(3))(7-4sqrt3)) = (49-16*3) = 1
Numerator 7+4sqrt(3) + 7-4sqrt(3) = 14
quadratic equation
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knight247
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For the 1st onerupsk wrote:Please help me in attached problems
x²-5x+6<0 and x²+5x+6=M
Lets solve the inequality first
x²-3x-2x+6<0
(x-3)(x-2)<0
Meaning both factors have opposing signs
(x-3)<0 and (x-2)>0
or
(x-3)>0 and (x-2)<0
We get two sets of values from the above
x<3 and x>2 from the first one. aka 2<x<3
from the second one we get
x>3 and x<2. This set of values is redundant as x can't be greater than 3 and less than 2.
So we consider 2<x<3. Now let us consider x=2 and x=3 as these exceed the extreme values. Substituting them separately in x²+5x+6=M we get
M=20 and M=30. Since we have considered x slightly greater than the extremes we have 20<M<30. Hence B
