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4 years ago

What are the roots of -12x2 + 11x − 3 = 0?

1 answer:

5 0

This quadratic does not have any real roots.
It is a parabola that opens downward with vertex at (11/24, -.48) which means the vertex is below the x-axis and therefore will not have any real roots.

Imaginary roots would be
x = [11 +- sqrt (121 - 4(-12)(-3)] / (-24)
x = [11 + - i sqrt(23) ] / (-24)

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anzhelika [568]

Well we know that... $a^{-1}=\frac{1}{a^{1} }\\$ .

So... $(\frac{1}{5})^{-1}=\frac{1}{\frac{1}{5}^{1}}=5\\\\$ .

Because $5^{3}=125$ ,

$\frac{1}{5}^{-1*3}=125\\ x=-1*3=-3$

answer: -3

8 0

4 years ago

Hii whats your name heehee

Roman55 [17]

Answer:

Kimberly

Step-by-step explanation:

My mother gave me the name

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2 years ago

A bird leaves its nest and travels 12 miles per hour downwind for x hours. On the return trip, the bird travels 2 miles per hour

elena-s [515]

3 hours

On the way there, 1.5x12 = 18
On the way back, 1.5x10 = 15

4 0

3 years ago

Read 2 more answers

Multiplicative inverse of a neggative<br>rational number is

lions [1.4K]

$for \: a \: rational \: number \: \frac{a}{b} \: such \: that \: 0 > a \: \: \: \: \: \: 0 < b \\ its \: multiplicative \: inverse \: is \: - \frac{b}{a}$

6 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago