Solved on Feb 29, 2024

Find the values of $x$ where $3x^2 - 19x - 14 > 0$ .

STEP 1

Assumptions
1. The function is given by $k(x) = 3x^2 - 19x - 14$ .
2. We need to find all values of $x$ for which $k(x) > 0$ .

STEP 2

To find the values of $x$ for which $k(x) > 0$ , we need to determine the roots of the quadratic equation $k(x) = 0$ . We can do this by using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 3$ , $b = -19$ , and $c = -14$ .

STEP 3

First, calculate the discriminant $(b^2 - 4ac)$ of the quadratic equation.
$\Delta = b^2 - 4ac$
$\Delta = (-19)^2 - 4(3)(-14)$

STEP 4

Calculate the value of the discriminant.
$\Delta = 361 + 168$ $\Delta = 529$

STEP 5

Since the discriminant $\Delta > 0$ , there are two distinct real roots. Now, use the quadratic formula to find the roots.
$x = \frac{-(-19) \pm \sqrt{529}}{2 \cdot 3}$

STEP 6

Simplify the expression inside the quadratic formula.
$x = \frac{19 \pm 23}{6}$

STEP 7

Calculate the two values of $x$ .
$x_1 = \frac{19 + 23}{6} = \frac{42}{6} = 7$ $x_2 = \frac{19 - 23}{6} = \frac{-4}{6} = -\frac{2}{3}$

STEP 8

The roots of the equation $k(x) = 0$ are $x = 7$ and $x = -\frac{2}{3}$ . The quadratic function $k(x)$ will be positive between the roots and negative outside the roots since the leading coefficient $a = 3$ is positive.

STEP 9

The values of $x$ for which $k(x) > 0$ are in the intervals $(-\infty, -\frac{2}{3})$ and $(7, \infty)$ .
The values of $x$ for which $k(x) > 0$ are all $x$ such that $x < -\frac{2}{3}$ or $x > 7$ .

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Find the values of $x$ where $3x^2 - 19x - 14 > 0$ .

STEP 1

Assumptions
1. The function is given by $k(x) = 3x^2 - 19x - 14$ .
2. We need to find all values of $x$ for which $k(x) > 0$ .

STEP 2

To find the values of $x$ for which $k(x) > 0$ , we need to determine the roots of the quadratic equation $k(x) = 0$ . We can do this by using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 3$ , $b = -19$ , and $c = -14$ .

STEP 3

First, calculate the discriminant $(b^2 - 4ac)$ of the quadratic equation.
$\Delta = b^2 - 4ac$
$\Delta = (-19)^2 - 4(3)(-14)$

STEP 4

Calculate the value of the discriminant.
$\Delta = 361 + 168$ $\Delta = 529$

STEP 5

Since the discriminant $\Delta > 0$ , there are two distinct real roots. Now, use the quadratic formula to find the roots.
$x = \frac{-(-19) \pm \sqrt{529}}{2 \cdot 3}$

STEP 6

Simplify the expression inside the quadratic formula.
$x = \frac{19 \pm 23}{6}$

STEP 7

Calculate the two values of $x$ .
$x_1 = \frac{19 + 23}{6} = \frac{42}{6} = 7$ $x_2 = \frac{19 - 23}{6} = \frac{-4}{6} = -\frac{2}{3}$

STEP 8

The roots of the equation $k(x) = 0$ are $x = 7$ and $x = -\frac{2}{3}$ . The quadratic function $k(x)$ will be positive between the roots and negative outside the roots since the leading coefficient $a = 3$ is positive.

STEP 9

The values of $x$ for which $k(x) > 0$ are in the intervals $(-\infty, -\frac{2}{3})$ and $(7, \infty)$ .
The values of $x$ for which $k(x) > 0$ are all $x$ such that $x < -\frac{2}{3}$ or $x > 7$ .

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Find the values of xxx where 3x2−19x−14>03x^2 - 19x - 14 > 03x2−19x−14>0.

STEP 1

Assumptions1. The function is given by k(x)=3x2−19x−14k(x) = 3x^2 - 19x - 14k(x)=3x2−19x−14.2. We need to find all values of xxx for which k(x)>0k(x) > 0k(x)>0.

STEP 2

STEP 3

First, calculate the discriminant (b2−4ac)(b^2 - 4ac)(b2−4ac) of the quadratic equation.Δ=b2−4ac\Delta = b^2 - 4acΔ=b2−4acΔ=(−19)2−4(3)(−14)\Delta = (-19)^2 - 4(3)(-14)Δ=(−19)2−4(3)(−14)

STEP 4

Calculate the value of the discriminant.Δ=361+168\Delta = 361 + 168Δ=361+168 Δ=529\Delta = 529Δ=529

STEP 5

Since the discriminant Δ>0\Delta > 0Δ>0, there are two distinct real roots. Now, use the quadratic formula to find the roots.x=−(−19)±5292⋅3x = \frac{-(-19) \pm \sqrt{529}}{2 \cdot 3}x=2⋅3−(−19)±529​​

STEP 6

Simplify the expression inside the quadratic formula.x=19±236x = \frac{19 \pm 23}{6}x=619±23​

STEP 7

Calculate the two values of xxx.x1=19+236=426=7x_1 = \frac{19 + 23}{6} = \frac{42}{6} = 7x1​=619+23​=642​=7 x2=19−236=−46=−23x_2 = \frac{19 - 23}{6} = \frac{-4}{6} = -\frac{2}{3}x2​=619−23​=6−4​=−32​

STEP 8

The roots of the equation k(x)=0k(x) = 0k(x)=0 are x=7x = 7x=7 and x=−23x = -\frac{2}{3}x=−32​. The quadratic function k(x)k(x)k(x) will be positive between the roots and negative outside the roots since the leading coefficient a=3a = 3a=3 is positive.

STEP 9

The values of xxx for which k(x)>0k(x) > 0k(x)>0 are in the intervals (−∞,−23)(-\infty, -\frac{2}{3})(−∞,−32​) and (7,∞)(7, \infty)(7,∞).The values of xxx for which k(x)>0k(x) > 0k(x)>0 are all xxx such that x<−23x < -\frac{2}{3}x<−32​ or x>7x > 7x>7.

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Find the values of $x$ where $3x^2 - 19x - 14 > 0$ .

Assumptions
1. The function is given by $k(x) = 3x^2 - 19x - 14$ .
2. We need to find all values of $x$ for which $k(x) > 0$ .

First, calculate the discriminant $(b^2 - 4ac)$ of the quadratic equation.
$\Delta = b^2 - 4ac$
$\Delta = (-19)^2 - 4(3)(-14)$

Calculate the value of the discriminant.
$\Delta = 361 + 168$ $\Delta = 529$

Since the discriminant $\Delta > 0$ , there are two distinct real roots. Now, use the quadratic formula to find the roots.
$x = \frac{-(-19) \pm \sqrt{529}}{2 \cdot 3}$

Simplify the expression inside the quadratic formula.
$x = \frac{19 \pm 23}{6}$

Calculate the two values of $x$ .
$x_1 = \frac{19 + 23}{6} = \frac{42}{6} = 7$ $x_2 = \frac{19 - 23}{6} = \frac{-4}{6} = -\frac{2}{3}$

The roots of the equation $k(x) = 0$ are $x = 7$ and $x = -\frac{2}{3}$ . The quadratic function $k(x)$ will be positive between the roots and negative outside the roots since the leading coefficient $a = 3$ is positive.

The values of $x$ for which $k(x) > 0$ are in the intervals $(-\infty, -\frac{2}{3})$ and $(7, \infty)$ .
The values of $x$ for which $k(x) > 0$ are all $x$ such that $x < -\frac{2}{3}$ or $x > 7$ .