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For how many integral values of x, the expression x^2 – 7x -2 is less

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shameekv1989
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For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink] New post   19 Apr 2020, 06:00
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For how many integral values of x, the expression x^2 – 7x -2 is always less than 6 and greater than -8?

A) 0
B) 1
C) 2
D) 3
E) 4
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C
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For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink] New post   19 Apr 2020, 06:44
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shameekv1989 wrote:
For how many integral values of x, the expression x^2 – 7x -2 is always less than 6 and greater than -8?

A) 0
B) 1
C) 2
D) 3
E) 4


x^2 – 7x -2 < 6
i.e. x^2 – 7x -8 < 0
i.e. x^2 – 8x + x -8 < 0
i.e. (x – 8)*(x + 1) < 0

i.e. -1 < x < 8


x^2 – 7x -2 > -8
i.e. x^2 – 7x + 6 > 0
i.e. x^2 – 6x - x + 6 > 0
i.e. (x – 6)*(x - 1) > 0

i.e. x > 6 or x < 1

COmbining the range

6 < x < 8 and -1 < x< 1

i.e. two integer values of x {7 and 0}

Answer: Option C
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Re: For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink] New post   13 Oct 2021, 07:26
\(-8 < X^2 - 7*X - 2 < 6\)

\(\frac{ -b +or- \sqrt{(b^2 - 4 * a * C)} }{ 2*a }\) = \(\frac{ 7 +or- \sqrt{57} }{ 2 }\)

\(= 3.5 +/- \sqrt{\frac{ 57}{4 }}\)

This is approximately \(3.5 +/- \sqrt{14}\)

14 is in between 9 & 16 nearer the bigger number 16. So, it must be slightly more than 3.5.

Now, let's take \(3.5 +/- 3.7\) and try to round that off to closest integers.

So, now we get integer values of X as 0 & 7.

Here, the most important catch is that we should not reject the value "7" looking at the condition less than 6. Remember, the condition doesn't say that X has to be less than 6. It says the expression " \(X^2 - 7*X - 2\) " has to be less than 6.

So, the answer is 2. (i.e. choice C)
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Re: For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink] New post   17 Oct 2021, 20:22
Hi in your solution (x-6)(x-1)>0
hence x>6 and x>1
but you have written x<1.
Am I missing something?
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Re: For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink] New post   17 Oct 2021, 20:36
shameekv1989 wrote:
For how many integral values of x, the expression x^2 – 7x -2 is always less than 6 and greater than -8?

A) 0
B) 1
C) 2
D) 3
E) 4



Solution-

x^2-7x-2<6
x^2-7x-8<0
x^2-8x+x-8<0
(x+1)(x-8)<0

so -1<x<8, x= 0,1,2,3,4,5,6,7 (8 values)

x^2-7x-2>-8
x^2-7x+6>0
x^2-6x-x+6>0
(x-1)(x-6)>0

x<1 and x>6

Values of X remain from above

x= 0, 7

so only 2 values of x will satisfy the conditions.
Answer is C.
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parasma
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Re: For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink] New post   17 Oct 2021, 20:41
Surgicalgoblin wrote:
Hi in your solution (x-6)(x-1)>0
hence x>6 and x>1
but you have written x<1.
Am I missing something?


Dear,

Wavy line method is used to find the range.
Solution is attached in photograph.
Attachments

IMG_20211018_090820.jpg
IMG_20211018_090820.jpg [ 1.49 MiB | Viewed 1387 times ]

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Re: For how many integral values of x, the expression x^2 – 7x -2 is less [#permalink]
17 Oct 2021, 20:41
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