How Do You Find the Y-Intercept and Solve Quadratics by Completing the Square?

[Richay]

A couple more I'm having problems with.

First with linear intercepts.

Find the y-intercept of a line going through (3, 0) and having a slope of -5

y=mx+b

m = -5

And from then on, I don't know how to continue with this problem.

------

Now, completing the square.

Solve: Solve: 4x^2 - x - 5 = 0

Steps:
4x^2 – x = 5

x^2 - 1/4x = 5/4

-1/4 --> 1/16

x^2 - 1/2x + 1/16 = 5/4 + 1/16

(x-1/4)^2 = 21/16

Solve for x now right? Well i think i already messed up in one of my earlier steps. I need help.

Thanks for any help...

 

[Pyrrhus]

Do you know what b means in y = mx +b?? (hint: that's what you are looking for)

Check again your work for completing the square.

 

[Richay]

Do you know what b means in y = mx +b?? (hint: that's what you are looking for)

Check again your work for completing the square.

b is the value of the y-intercept.

 

[Curious3141]

A couple more I'm having problems with.

First with linear intercepts.

Find the y-intercept of a line going through (3, 0) and having a slope of -5

y=mx+b

m = -5

And from then on, I don't know how to continue with this problem.

The y-intercept is the value y assumes when x = 0. This is when the line cuts through the y-axis. In the equation y = mx + b, what does y equal when x = 0 ?

You're given a point on the line which is essentially the y-value for a specific x-value. You also know the slope (m). Now put those back in and work out b.
------

Now, completing the square.

Solve: Solve: 4x^2 - x - 5 = 0

Steps:
4x^2 – x = 5

x^2 - 1/4x = 5/4

-1/4 --> 1/16

x^2 - 1/2x + 1/16 = 5/4 + 1/16

(x-1/4)^2 = 21/16

Solve for x now right? Well i think i already messed up in one of my earlier steps. I need help.

Thanks for any help...

It's OK up to this step : x^2 - 1/4x = 5/4

In completing the square, you are basically comparing the left hand side (LHS) to (x-c)^2 = x^2 - 2cx + c^2.

Now compare the coefficient of the x term in that expansion to your own LHS and find out the value of c. The square of that expression then bears a remarkable resemblance to your original equation, and all you need to do is add constants to make it identical, then you can solve it.