Solve 3x to the second power -2x-8=0 using an algebraic method

Mathematics

1 answer:

MatroZZZ [7]2 years ago

7 0

Answer:

$x=2\textrm{ or } x=-\frac{4}{3}$

Step-by-step explanation:

Given:

The equation is given as:

$3x^{2}-2x-8=0\\3x^{2}-6x+4x-8=0\\3x(x-2)+4(x-2)=0\\(x-2)(3x+4)=0\\\\x-2=0\textrm{ or }3x+4=0\\x-2+2=0+2 \textrm{ or }3x+4-4=0-4\\x=2\textrm{ or }3x=-4\\x=2\textrm{ or}\\x=-\frac{4}{3}$

Therefore, the possible values of $x$ are 2 and $-\frac{4}{3}$

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Consider the line x+5y=6

Tems11 [23]

Answer:

$x + 5y = 6$ is perpendicular to $y = 5x + \frac{6}{5}$ and parallel to $y = -\frac{1}{5}x + 2\\$

Step-by-step explanation:

First, convert the equation to standard form so that y is isolated.

x + 5y = 6 --> x - 6 = -5y (divide both sides by -5) --> $y = -\frac{1}{5}x + \frac{6}{5}$

A perpendicular line will have a slope that is the opposite reciprocal of the original slope (meaning you flip the numerator and denominator then make it negative).

$-\frac{1}{5}$ is perpendicular to $-(\frac{-5}{1} )$ which simplifies to 5.

A parallel line will have the same slope, but the y-intercept will be different. It can be pretty much any number as long as the original slope is used in the new equation.

$y = -\frac{1}{5}x + \frac{6}{5}\\$ is parallel to $y = -\frac{1}{5}x + 2\\$ just like $y = -\frac{1}{5}x - \frac{100}{23}$ .