finding equation for straight line; solving radical equation

[kjv4]

I was just wondering if someone could help me with a few pre cal problems.
1. Find and equation of the line that passes through the x-intercept [(-1/3),0] and the y-intercept [0, (-8/3)]

and

2.solve for x: (square root) 15x+4 = negative (square root) 2x+3

 

[Deleted member 4993]

Re: Pre Cal!!

I was just wondering if someone could help me with a few pre cal problems.
1. Find and equation of the line that passes through the x-intercept [(-1/3),0] and the y-intercept [0, (-8/3)]

If the x-intercept length is = a and the y-intercept length is = b, then the equation of the line is:

x/a + y/b = 1

and

2.solve for x: (square root) 15x+4 = negative (square root) 2x+3

Is your problem

\(\displaystyle sqrt{15x} + 4 = - sqrt{2x} + 3\)

 

[jwpaine]

For number 1: first you need to find the slope of the line that passes through both points (x1,y1) and (x2,y2)

slope = m = \(\displaystyle \L \frac{y_2 - y_1}{x_2 - x_1}\)

And then use the point-slope form of the equation for a line:

\(\displaystyle \L y - y_1 = m(x - x_1)\)

That will give you the equation for the line through both of your points.



For number 2: solve:
\(\displaystyle \L \sqrt{15x + 4} = -\sqrt{2x + 3}\)

square both sides because \(\displaystyle \L \sqrt{x} \cdot \sqrt{x} = x\)

\(\displaystyle \L \sqrt{15x + 4}^{2} = -\sqrt{2x + 3}^{2}\)

\(\displaystyle \L 15x + 4 = 2x + 3\) (negative squared = positive)

now combine like terms / simplify and solve.

John.

 

[morson]

If jw interpreted the equation correctly, what do you notice about it?

The left side is a principal square root. It is always positive or 0. The right side is a negative principal square root. It is always negative or 0. From this information, there can only be a solution if there exists a number x such that the expressions inside both square roots are 0. If you follow on from what jw did, you'll come across what's called an extraneous solution.