If 7<x^2-4x+12, which of the following MUST be true?

If \(7<x^2-4x+12\), which of the following MUST be true?

\(7<x^2-4x+12.....x^2-4x+5>0....X^2-5X+X>0.....(X-2)^2>-1\)....This is true for all values of x.....

So x could be anything

i) \(3^{x-2}>0\)...True for all values of x

ii) \(|x^3+1|>0\)...True for all values except when x=-1

iii) \(\sqrt{(x+2)^2}>0\).....True for all values except when x=-2
_________________

Simplifying the stem x^2- 4x + 5>0
---> (x-2)^2 + 1 >0 ---> something positive + 1 will always be greater than zero. Hence x can take any value.

1. Does not matter what value x takes. Always true.
2. Not true always . Plug in x=-1 . Inequality will not hold 0 is not greater than 0.
3. Not always true . Plug in x=-2. Inequality will not hold.
A
_________________

APPROACH #1: Start with the given inequality and then check the statements
Given: \(7<x^2-4x+12\)
Subtract 8 from both sides to get: \(-1<x^2-4x+4\)
Factor right side to get: \(-1<(x-2)^2\)
Since \((x-2)^2\) is ALWAYS greater than or equal to zero, we can see that the inequality is true for ALL values of x.
In other words, x can equal ANY number

i) \(3^{x-2}>0\)
This is true for all values of x.
So, statement i is MUST be true

ii) \(|x^3+1|>0\)
This inequality is NOT satisfied when \(x = -1\)
Since x can equal -1, statement ii need not be true

iii) \(\sqrt{(x+2)^2}>0\)
This inequality is NOT satisfied when \(x = -2\)
Since x can equal -2, statement iii need not be true

Answer: A
--------------------------------------

APPROACH #2: Start with the statements and then check the given inequality

i) \(3^{x-2}>0\)
This is true for all values of x.
So, statement i is MUST be true

ii) \(|x^3+1|>0\)
This inequality is NOT satisfied when \(x = -1\)
Check the given inequality to see if x CAN equal -1
Plug in x= -1 to get: \(7<(-1)^2-4(-1)+12\)
Simplify: \(7<17\). WORKS.
Since x can equal -1, statement ii need not be true

iii) \(\sqrt{(x+2)^2}>0\)
This inequality is NOT satisfied when \(x = -2\)
Plug in x= -2 to get: \(7<(-2)^2-4(-2)+12\)
Simplify: \(7<24\). WORKS.
Since x can equal -2, statement iii need not be true

Answer: A


Cheers,
Brent

What a beautifully hard question.

(1st) re-express the Given expression


(X)^2 - 4X + 12 > 7

((X)^2 - 4X + 4) + 8 > 7

(X - 2)^2 > (-)1

Since the output of any real value Squared will always be 0 or (+)positive, every real value of X on the number line will satisfy the given inequality.

Therefore, when looking at the given expressions in the Roman numerals, the only statement that MUST necessarily be true is the statement that allows X to take any real value on the number line.


-I-

(3)^(x - 2) > 0

Rule: even if the exponent is negative, the base of the exponential term never changes - it alway stays the same.

Therefore, the given term on the left hand side will always be positive for any real value of X that you plug in.

I must be true always.


-II-
[(x)^3 + 1] > 0

From the given, we know that X can be any real value on the number line.

However, if we plug X = (-1) into Roman numeral II, the inequality will not hold. 0 = 0

So for the value of X = (-1) this statement does not have to be true


-III-
Using the same logic as above, for the value of X = (-2) ——- which is the value that makes the expression on the left hand side of this inequality equal to 0——- the inequality will not hold because 0 = 0


(A)only Roman numeral I must be true in all circumstances.

Posted from my mobile device

This question can best be solved by putting values for x in 7<x2−4x+127<x2−4x+12

x can have any value from 2 to - infinitey

i) 3^x−2>0 --> must always be true

ii) |x3+1|>0 --> when x = -1 the result is 0

iii)when x = -2, the result 0

hence 1) is the only option which must be true