CALCU-how to do question 4)g (x+4)^2 25e (x+3)(1-

how to do question 4)

Solution by Steps

step 1

To solve the inequality \frac{5}{2x^2 + x - 15} < 0 for

x

, we first need to find the roots of the quadratic equation

2x^2 + x - 15 = 0

step 2

Using the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

, where

a = 2

,

b = 1

, and

c = -15

, we get:

x = \frac{-1 \pm \sqrt{1 + 120}}{4} = \frac{-1 \pm \sqrt{121}}{4} = \frac{-1 \pm 11}{4}

Thus, the roots are

x = \frac{10}{4} = \frac{5}{2}

and

x = \frac{-12}{4} = -3

step 3

The quadratic

2x^2 + x - 15

can be factored as

2(x + 3)(x - \frac{5}{2})

step 4

To determine where \frac{5}{2x^2 + x - 15} < 0, we analyze the sign of the quadratic expression

2(x + 3)(x - \frac{5}{2})

step 5

The quadratic expression changes sign at the roots

x = -3

and

x = \frac{5}{2}

. We test intervals around these points: - For x < -3, both factors

(x + 3)

and

(x - \frac{5}{2})

are negative, so the product is positive. - For -3 < x < \frac{5}{2},

(x + 3)

is positive and

(x - \frac{5}{2})

is negative, so the product is negative. - For x > \frac{5}{2}, both factors are positive, so the product is positive

step 6

Therefore, \frac{5}{2x^2 + x - 15} < 0 when -3 < x < \frac{5}{2}

Answer

-3 < x < \frac{5}{2}

Key Concept

Solving quadratic inequalities

Explanation

To solve a quadratic inequality, find the roots of the quadratic equation, factorize it, and analyze the sign of the expression in different intervals around the roots.

how to do 5b)

Solution by Steps

step 1

To solve the inequality

x^2 + 4x - 21 \leq 0

, we first need to find the roots of the quadratic equation

x^2 + 4x - 21 = 0

step 2

Using the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

, where

a = 1

,

b = 4

, and

c = -21

, we get:

x = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}

step 3

This gives us the roots

x = 3

and

x = -7

. Therefore, the inequality

x^2 + 4x - 21 \leq 0

holds for

-7 \leq x \leq 3

step 4

Next, we solve the inequality x^2 - 9x + 8 > 0. We find the roots of the quadratic equation

x^2 - 9x + 8 = 0

using the quadratic formula:

x = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2}

step 5

This gives us the roots

x = 8

and

x = 1

. Therefore, the inequality x^2 - 9x + 8 > 0 holds for x < 1 or x > 8

step 6

Combining the two inequalities, we need to find the intersection of the intervals

-7 \leq x \leq 3

and x < 1 or x > 8. The intersection is -7 \leq x < 1

Answer

-7 \leq x < 1

Key Concept

Solving quadratic inequalities

Explanation

To solve quadratic inequalities, we find the roots of the quadratic equations and determine the intervals where the inequalities hold. Then, we find the intersection of these intervals.

HOW TO DO 5C)

Solution by Steps

step 1

To solve the inequality x^2 + x - 2 > 0, we first need to find the roots of the quadratic equation

x^2 + x - 2 = 0

step 2

Using the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

, where

a = 1

,

b = 1

, and

c = -2

, we get:

x = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}

step 3

This gives us the roots

x = 1

and

x = -2

step 4

The quadratic

x^2 + x - 2

is positive outside the interval

(-2, 1)

. Therefore, x^2 + x - 2 > 0 for x < -2 or x > 1

step 5

Next, we solve the inequality

x^2 - 2x - 3 \geq 0

step 6

Using the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

, where

a = 1

,

b = -2

, and

c = -3

, we get:

x = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}

step 7

This gives us the roots

x = 3

and

x = -1

step 8

The quadratic

x^2 - 2x - 3

is non-negative outside the interval

(-1, 3)

. Therefore,

x^2 - 2x - 3 \geq 0

for

x \leq -1

or

x \geq 3

step 9

The solution to both inequalities is the intersection of the intervals: (x < -2 \text{ or } x > 1) \cap (x \leq -1 \text{ or } x \geq 3)

step 10

This simplifies to: x < -2 \text{ or } x \geq 3

Answer

x < -2 or

x \geq 3

Key Concept

Solving quadratic inequalities

Explanation

To solve quadratic inequalities, find the roots of the quadratic equations and determine the intervals where the quadratic expressions are positive or non-negative. The solution is the intersection of these intervals.

how to do question 6

Solution by Steps

step 1

We start with the inequality 2^{x^2 - 3x - 40} > 1

step 2

Since 2^y > 1 for any y > 0, we need to solve x^2 - 3x - 40 > 0

step 3

Factor the quadratic expression:

x^2 - 3x - 40 = (x - 8)(x + 5)

step 4

Set the factors greater than 0: (x - 8)(x + 5) > 0

step 5

Determine the critical points:

x = 8

and

x = -5

step 6

Analyze the intervals determined by the critical points:

(-\infty, -5)

,

(-5, 8)

, and

(8, \infty)

step 7

Test points in each interval to determine where the product is positive: - For x < -5, choose

x = -6

: (-6 - 8)(-6 + 5) = (-14)(-1) > 0. - For -5 < x < 8, choose

x = 0

: (0 - 8)(0 + 5) = (-8)(5) < 0. - For x > 8, choose

x = 9

: (9 - 8)(9 + 5) = (1)(14) > 0

step 8

Therefore, the solution to the inequality is x < -5 or x > 8

Answer

x < -5 or x > 8

Key Concept

Solving quadratic inequalities

Explanation

To solve the inequality 2^{x^2 - 3x - 40} > 1, we need to find where the quadratic expression

x^2 - 3x - 40

is greater than 0. This involves factoring the quadratic, finding the critical points, and testing intervals to determine where the product of the factors is positive.

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