Find the range of values of x that satisfy the inequality (x + 1)(x –

Hey Everyone,

Please find below the solution of the given problem :

Solution:

Rewriting the inequality to easily identify the zero points
\((x + 1) (x – 2) > 4\)

\(x^2 – x – 2 – 4 > 0\)

\(x^2 – x – 6 > 0\)

\((x – 3) (x+2) > 0\)

Plotting the zero points and drawing the wavy line:



Required Range: x > 3 or x < -2

Correct Answer: Option E


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Please post the options also.

This is what I have calculated.

(x+1)(x-2) > 4

or x^2 -x -2 > 4

or x^2 -x -6 > 0

or (x+2)(x-3)>0

or while drawing on the number line, will get the range of x as (-infinity, -2) U ( 3,infinity)

Please correct me if I missed anything.
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(x+1)(x-2)>4
(x-3)(x+2)>0 rearrange the inequality
So zero points are 3 & -2
and -2< x>3 to hold the equation true
I hope my answer is logic

\((x + 1) (x – 2) > 4\) Only when x > 4

So, There can be infinite numbers which are > 4..

Hence I am also with the answer as infinite..
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First of all, ONLY x>4 is not correct, even if you input x=3.1 which is slightly greater than 3, the inequality yields value more than 4.
Secondly, for values lesser than -2 also, inequality yields value greater than 4. example: Put x= -3 (which is lesser than -2), inequality gives value more than 4.
However, it is recommended to put the inequality in standard format like <0, >0, <= 0 [(x+1)(x-2)-4>0 ]etc. In such case solving the inequality will be easier..

x should be lesser than -2 which is -2>x. In your equation it's written opposite.

(x+1)(x–2)>4
or,x^2–x–2–4>0
or,x^2–x–6>0
or,(x–3)(x+2)>0
or,x > 3 or x < -2

correct answer E

if (x – 3) (x+2) > 0 then both terms are greater than zero, right? then how we get x<-2? Please help me to understand it clearly. i am confused.

Find the range of values of x that satisfy the inequality (x+1)(x–2)>4

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thank you so much....i have substitute and check

i have the exact same question

\((x + 1)(x – 2) > 4\)

\(x^2 - 2x + x - 2 > 4\)

\((x - 3) (x + 2)>0\)

Here is how to solve the above inequality easily. The "roots", in ascending order, are -2, and 3, which gives us 3 ranges:

\(x < -2\);

\(-2 < x < 3\);

\(3 < x\).

Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 10, then all tow multiples will be positive, giving a positive result for the whole expression. So when \(x > 3\), the expression is positive. Now the trick: as in the 3rd range, the expression is positive, then in the 2nd it'll be negative, and finally in the 1st, it'll be positive: \(\text{(+ - +)}\). So, the ranges when the expression is positive are: \(x < -2\) and \(x > 3\).

P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative are: \(-2 < x < 4\).
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