Solved on Jan 17, 2024

Find the upper bound of the inequality $y \leq x^2 + 6x + 4$ .

STEP 1

Assumptions
1. We are given the inequality $y \leq x^{2}+6x+4$ .
2. We need to find the solution set for this inequality.

STEP 2

The given inequality is a quadratic inequality in the form $y \leq ax^{2} + bx + c$ , where $a = 1$ , $b = 6$ , and $c = 4$ .

STEP 3

To solve the inequality, we first consider the corresponding quadratic equation $x^{2} + 6x + 4 = 0$ .

STEP 4

We can solve the quadratic equation by factoring, completing the square, or using the quadratic formula. We will use the quadratic formula here.

STEP 5

The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

STEP 6

Plug in the values of $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-6 \pm \sqrt{6^{2} - 4(1)(4)}}{2(1)}$

STEP 7

Calculate the discriminant $b^{2} - 4ac$ .
$\Delta = 6^{2} - 4(1)(4) = 36 - 16 = 20$

STEP 8

Since the discriminant is positive, we have two distinct real roots. Now, compute the roots using the quadratic formula.
$x = \frac{-6 \pm \sqrt{20}}{2}$

STEP 9

Simplify the square root of the discriminant.
$\sqrt{20} = 2\sqrt{5}$

STEP 10

Substitute the simplified square root back into the formula.
$x = \frac{-6 \pm 2\sqrt{5}}{2}$

STEP 11

Divide the terms in the numerator by the denominator.
$x = -3 \pm \sqrt{5}$

STEP 12

Now we have the two roots of the equation $x^{2} + 6x + 4 = 0$ , which are $x = -3 + \sqrt{5}$ and $x = -3 - \sqrt{5}$ .

STEP 13

The roots divide the number line into intervals. We need to determine the sign of the quadratic expression $x^{2} + 6x + 4$ within each interval.

STEP 14

Choose test points from each interval, which are to the left of $x = -3 - \sqrt{5}$ , between $x = -3 - \sqrt{5}$ and $x = -3 + \sqrt{5}$ , and to the right of $x = -3 + \sqrt{5}$ .

STEP 15

Evaluate the quadratic expression at each test point to determine the sign of the expression in each interval.

STEP 16

Since the leading coefficient $a = 1$ is positive, the parabola opens upwards, and the expression will be negative between the roots and positive outside of the roots.

STEP 17

The inequality $y \leq x^{2}+6x+4$ means we are looking for the values of $x$ for which the quadratic expression is less than or equal to zero.

STEP 18

The solution set for the inequality includes all $x$ values between the roots, including the roots themselves.

STEP 19

Write the solution set in interval notation.
$x \in [-3 - \sqrt{5}, -3 + \sqrt{5}]$

STEP 20

The final solution to the inequality $y \leq x^{2}+6x+4$ is that $y$ is less than or equal to any value of $x$ within the interval $[-3 - \sqrt{5}, -3 + \sqrt{5}]$ .

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Find the upper bound of the inequality y≤x2+6x+4y \leq x^2 + 6x + 4y≤x2+6x+4.

STEP 1

Assumptions1. We are given the inequality y≤x2+6x+4y \leq x^{2}+6x+4y≤x2+6x+4.2. We need to find the solution set for this inequality.

STEP 2

The given inequality is a quadratic inequality in the form y≤ax2+bx+cy \leq ax^{2} + bx + cy≤ax2+bx+c, where a=1a = 1a=1, b=6b = 6b=6, and c=4c = 4c=4.

STEP 3

To solve the inequality, we first consider the corresponding quadratic equation x2+6x+4=0x^{2} + 6x + 4 = 0x2+6x+4=0.

STEP 4

We can solve the quadratic equation by factoring, completing the square, or using the quadratic formula. We will use the quadratic formula here.

STEP 5

The quadratic formula is given by:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}x=2a−b±b2−4ac​​

STEP 6

Plug in the values of aaa, bbb, and ccc into the quadratic formula.x=−6±62−4(1)(4)2(1)x = \frac{-6 \pm \sqrt{6^{2} - 4(1)(4)}}{2(1)}x=2(1)−6±62−4(1)(4)​​

STEP 7

Calculate the discriminant b2−4acb^{2} - 4acb2−4ac.Δ=62−4(1)(4)=36−16=20\Delta = 6^{2} - 4(1)(4) = 36 - 16 = 20Δ=62−4(1)(4)=36−16=20

STEP 8

Since the discriminant is positive, we have two distinct real roots. Now, compute the roots using the quadratic formula.x=−6±202x = \frac{-6 \pm \sqrt{20}}{2}x=2−6±20​​

STEP 9

Simplify the square root of the discriminant.20=25\sqrt{20} = 2\sqrt{5}20​=25​

STEP 10

Substitute the simplified square root back into the formula.x=−6±252x = \frac{-6 \pm 2\sqrt{5}}{2}x=2−6±25​​

STEP 11

Divide the terms in the numerator by the denominator.x=−3±5x = -3 \pm \sqrt{5}x=−3±5​

STEP 12

Now we have the two roots of the equation x2+6x+4=0x^{2} + 6x + 4 = 0x2+6x+4=0, which are x=−3+5x = -3 + \sqrt{5}x=−3+5​ and x=−3−5x = -3 - \sqrt{5}x=−3−5​.

STEP 13

The roots divide the number line into intervals. We need to determine the sign of the quadratic expression x2+6x+4x^{2} + 6x + 4x2+6x+4 within each interval.

STEP 14

Choose test points from each interval, which are to the left of x=−3−5x = -3 - \sqrt{5}x=−3−5​, between x=−3−5x = -3 - \sqrt{5}x=−3−5​ and x=−3+5x = -3 + \sqrt{5}x=−3+5​, and to the right of x=−3+5x = -3 + \sqrt{5}x=−3+5​.

STEP 15

Evaluate the quadratic expression at each test point to determine the sign of the expression in each interval.

STEP 16

Since the leading coefficient a=1a = 1a=1 is positive, the parabola opens upwards, and the expression will be negative between the roots and positive outside of the roots.

STEP 17

The inequality y≤x2+6x+4y \leq x^{2}+6x+4y≤x2+6x+4 means we are looking for the values of xxx for which the quadratic expression is less than or equal to zero.

STEP 18

The solution set for the inequality includes all xxx values between the roots, including the roots themselves.

STEP 19

Write the solution set in interval notation.x∈[−3−5,−3+5]x \in [-3 - \sqrt{5}, -3 + \sqrt{5}]x∈[−3−5​,−3+5​]

STEP 20

The final solution to the inequality y≤x2+6x+4y \leq x^{2}+6x+4y≤x2+6x+4 is that yyy is less than or equal to any value of xxx within the interval [−3−5,−3+5][-3 - \sqrt{5}, -3 + \sqrt{5}][−3−5​,−3+5​].

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Find the upper bound of the inequality $y \leq x^2 + 6x + 4$ .

Assumptions
1. We are given the inequality $y \leq x^{2}+6x+4$ .
2. We need to find the solution set for this inequality.

The given inequality is a quadratic inequality in the form $y \leq ax^{2} + bx + c$ , where $a = 1$ , $b = 6$ , and $c = 4$ .

To solve the inequality, we first consider the corresponding quadratic equation $x^{2} + 6x + 4 = 0$ .

The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Plug in the values of $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-6 \pm \sqrt{6^{2} - 4(1)(4)}}{2(1)}$

Calculate the discriminant $b^{2} - 4ac$ .
$\Delta = 6^{2} - 4(1)(4) = 36 - 16 = 20$

Since the discriminant is positive, we have two distinct real roots. Now, compute the roots using the quadratic formula.
$x = \frac{-6 \pm \sqrt{20}}{2}$

Simplify the square root of the discriminant.
$\sqrt{20} = 2\sqrt{5}$

Substitute the simplified square root back into the formula.
$x = \frac{-6 \pm 2\sqrt{5}}{2}$

Divide the terms in the numerator by the denominator.
$x = -3 \pm \sqrt{5}$

Now we have the two roots of the equation $x^{2} + 6x + 4 = 0$ , which are $x = -3 + \sqrt{5}$ and $x = -3 - \sqrt{5}$ .

The roots divide the number line into intervals. We need to determine the sign of the quadratic expression $x^{2} + 6x + 4$ within each interval.

Choose test points from each interval, which are to the left of $x = -3 - \sqrt{5}$ , between $x = -3 - \sqrt{5}$ and $x = -3 + \sqrt{5}$ , and to the right of $x = -3 + \sqrt{5}$ .

Since the leading coefficient $a = 1$ is positive, the parabola opens upwards, and the expression will be negative between the roots and positive outside of the roots.

The inequality $y \leq x^{2}+6x+4$ means we are looking for the values of $x$ for which the quadratic expression is less than or equal to zero.

The solution set for the inequality includes all $x$ values between the roots, including the roots themselves.

Write the solution set in interval notation.
$x \in [-3 - \sqrt{5}, -3 + \sqrt{5}]$

The final solution to the inequality $y \leq x^{2}+6x+4$ is that $y$ is less than or equal to any value of $x$ within the interval $[-3 - \sqrt{5}, -3 + \sqrt{5}]$ .