Find the number of integral values of x that satisfy|x2 + 3x - 1|

Question

Find the number of integral values of x that satisfy|x2 + 3x - 1| < 2|x| + 5Correct answer is '7'. Can you explain this answer?

Answer

|x² + 3x - 1| < 2|x| + 5

We have to find out the boundary points for each modulus and adjust the + or - sign after removing the modulus.

For the left expression, x² + 3x - 1y = x, the roots are

√13 ≈ 3.6
So, the approx roots are -3.3 and +0.3.

For the right side of the expression, the boundary point is 0

Let us denote the same in the number line.

For R1,

|x² + 3x - 1| < 2|x| + 5

x² + 3x - 1 < 2x + 5

x²+ x - 6 < 0

(x-2) (x +3) < 0

x ∈ (-3,2)

Overlapping region is

For R2,

|x² + 3x - 1| < 2|x| + 5

-x² - 3x + 1 < 2x + 5

-x² - 5x - 4 < 0

x² + 5x + 4 > 0

(x + 1) (x + 4) > 0

x ∈ R −(−4,−1)

Overlapping region is

For R3,

|x² + 3x - 1| < 2|x| + 5

-x² - 3x + 1 < - 2x + 5

- x² - x - 4 < 0

- x² + x + 4 > 0
This is possible for all values of x. Therefore the overlapping region is

For R4,

|x² + 3x - 1| < 2|x| + 5

-x² - 3x + 1 < - 2x + 5

x² + 5x - 6 < 0

(x + 6) (x -1) < 0

x ∈ (−6, 1)

Hence, the overall overlapping region is

Integral values are {-5, -4, -3, -2, -1, 0, 1}

Answer

To find the number of integral values of x that satisfy the given expression, we need to determine when the expression inside the absolute value brackets equals 0.

Setting x^2 + 3x - 1 = 0, we can solve this quadratic equation using factoring or the quadratic formula:

(x + 1)(x - 1) = 0

This equation is satisfied when x = -1 or x = 1.

Therefore, there are 2 integral values of x that satisfy the given expression.

Answer

|x² + 3x - 1| < 2|x| + 5

We have to find out the boundary points for each modulus and adjust the + or - sign after removing the modulus.

For the left expression, x² + 3x - 1y = x, the roots are

√13 ≈ 3.6
So, the approx roots are -3.3 and +0.3.

For the right side of the expression, the boundary point is 0

Let us denote the same in the number line.

For R1,

|x² + 3x - 1| < 2|x| + 5

x² + 3x - 1 < 2x + 5

x²+ x - 6 < 0

(x-2) (x +3) < 0

x ∈ (-3,2)

Overlapping region is

For R2,

|x² + 3x - 1| < 2|x| + 5

-x² - 3x + 1 < 2x + 5

-x² - 5x - 4 < 0

x² + 5x + 4 > 0

(x + 1) (x + 4) > 0

x ∈ R −(−4,−1)

Overlapping region is

For R3,

|x² + 3x - 1| < 2|x| + 5

-x² - 3x + 1 < - 2x + 5

- x² - x - 4 < 0

- x² + x + 4 > 0
This is possible for all values of x. Therefore the overlapping region is

For R4,

|x² + 3x - 1| < 2|x| + 5

-x² - 3x + 1 < - 2x + 5

x² + 5x - 6 < 0

(x + 6) (x -1) < 0

x ∈ (−6, 1)

Hence, the overall overlapping region is

Integral values are {-5, -4, -3, -2, -1, 0, 1}

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