Use discriminant for linear and quad. function.

Evangeline101 · May 21, 2016

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Back again :) Is this correct?

Thanks.

Mastermind01 · May 21, 2016

Again you are correct.

fresh_42 · May 21, 2016

Yes. Only that I have ##(-1.09 ; 1.73)## and ##(2.76 ; 13.27)##. But the path is correct.
It's always better to have exact numbers until the end and calculate the roots in a final step.

Ray Vickson · May 21, 2016

Evangeline101 said: ↑

1. The problem statement, all variables and given/known data
View attachment 101055

2. Relevant equations
View attachment 101057

View attachment 101061
3. The attempt at a solution
View attachment 101062
View attachment 101063

Back again :) Is this correct?

Thanks.

In thread after thread you ask "is this correct?...". That is a bad habit: as a matter of standard operating procedure you should check your answers for yourself (as you would need to do in an exam setting). Then, if you find something does not "work", you will know you have made an error somewhere, and asking for help then makes sense.

Mastermind01 · May 22, 2016

Ray Vickson said: ↑

In thread after thread you ask "is this correct?...". That is a bad habit: as a matter of standard operating procedure you should check your answers for yourself (as you would need to do in an exam setting). Then, if you find something does not "work", you will know you have made an error somewhere, and asking for help then makes sense.

I agree , I have already given the OP a way to check her work in a related similar thread.

Evangeline101 · May 22, 2016

Ray Vickson said: ↑

In thread after thread you ask "is this correct?...". That is a bad habit: as a matter of standard operating procedure you should check your answers for yourself (as you would need to do in an exam setting). Then, if you find something does not "work", you will know you have made an error somewhere, and asking for help then makes sense.

I understand, but can you suggest a less time consuming method than graphing, if I am doing an exam I won't have time to graph anything to check my answer. For this specific question, is there a way I can put the points back into the equation to verify, well maybe in this case I would have to graph since they are points of intersection.

Also I should be more clear when asking my question ( I will make sure to do that in future questions), in this case I would like to know if my process for part b is correct, does it make sense to use the linear equation to solve for y-values, and if so, is it because the linear equation is set equal to y?

Ray Vickson · May 22, 2016

Evangeline101 said: ↑

I understand, but can you suggest a less time consuming method than graphing, if I am doing an exam I won't have time to graph anything to check my answer. For this specific question, is there a way I can put the points back into the equation to verify, well maybe in this case I would have to graph since they are points of intersection.

Also I should be more clear when asking my question ( I will make sure to do that in future questions), in this case I would like to know if my process for part b is correct, does it make sense to use the linear equation to solve for y-values, and if so, is it because the linear equation is set equal to y?

To save time, don't graph anything; just check if your (x,y) values satisfy the requirement that they lie on both the line and the curve. That is easy: just plug in the numbers and calculate.

I have not checked your calculations for part (b), but I see that you have used the correct formulas. Again, you can check the solutions for yourself, by plugging in the numerical values and doing the calculations to see if everything works as it should.

Again: this is all just standard practice that you should get in the habit of doing.

Mark44 · May 22, 2016

Evangeline101 said: ↑

I understand, but can you suggest a less time consuming method than graphing, if I am doing an exam I won't have time to graph anything to check my answer.

Sure. You can substitute the points you found into each equation. If the point is on the graph, the equation should be a true statement.
In your work above, you found one solution to be ##x = \frac{5 + \sqrt{133}}{6}##. The corresponding y value is ##y = \frac{15 + \sqrt{133}}{2}## If you substitute this x value into the right sides of the equations ##y = 3x^2 - 2x - 4## and ##y = 3x + 5## you should get the y value I show above. Note that the values I show are the exact values. If you approximate the x values using a calculator, the y values you get will be only as precise as the rounding you do.

Evangeline101 said:

For this specific question, is there a way I can put the points back into the equation to verify, well maybe in this case I would have to graph since they are points of intersection.

Also I should be more clear when asking my question ( I will make sure to do that in future questions), in this case I would like to know if my process for part b is correct, does it make sense to use the linear equation to solve for y-values, and if so, is it because the linear equation is set equal to y?

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Use discriminant for linear and quad. function.

Evangeline101

Attached Files:

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Mastermind01

fresh_42

Staff: Mentor

Ray Vickson

Mastermind01

Evangeline101

Ray Vickson

Mark44

Staff: Mentor

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