Solving quadratics using zeros

[mercuryv8]

So I have the equation y=x^2 - 2x-5

I graphicaly determine the zeros to be 3.45 and -1.45...shouldent I be able to substitute these values in for x and solve the equation...which should be zero.

3.45 gives and answer near zero 0.0025 (perhaps rounding error)

but -1.45 is nowhere close.

What's the deal?

Nic

[blike]

Both answers are correct -- rounding gives you near zeros.

 

For your second x value:

 

y = x^2 - 2x - 5

 

substituting -1.45

 

y = (-1.45)^2 - (-2.9) - 5 = 0

becomes

y = 2.10 + 2.9 - 5

 

y = 0

[the tree]

So I have the equation y=x2 - 2x-5

There's a typo there right? Maybe you meant y=x²-2x-5 ?

 

You might do better to use the quadratic formula for an exact answer.

For 0=ax²+bx+c, [math]x=\frac{-b\pm \sqrt{b^2 -4ac}}{2a}[/math].

[mercuryv8]

Both answers are correct -- rounding gives you near zeros.

 

For your second x value:

 

y = x^2 - 2x - 5

 

substituting -1.45

 

y = (-1.45)^2 - (-2.9) - 5 = 0

becomes

y = 2.10 + 2.9 - 5

 

y = 0

 

Yeah I messed up when I was solving the equation...subtracting a negative number.

 

Shouldn’t be doing quadratic work before I have a handle on adding and subtracting integers eh!

 

Monday is my excuse!

 

Thanks

 

Yes typo in my first post...sorry

[math]

y=x^2-2x-5

[/math]

 

Nic

[Dave]

Yeah I messed up when I was solving the equation...subtracting a negative number.

 

Shouldn’t be doing quadratic work before I have a handle on adding and subtracting integers eh!

 

I shouldn't worry about it. Most of the people I now know on the fourth year of a maths degree can't add up any more - I know I certainly can't