Finding the x- intercept

I would just like a second opinion please to ensure I am doing this correctly.

When y = 0

0 = -x^2 + 4x + 5
- 5 = 4x

x = - 1.25

Quote:

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Originally Posted by David Green

I would just like a second opinion please to ensure I am doing this correctly.

When y = 0

0 = -x^2 + 4x + 5
- 5 = 4x

x = - 1.25

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The x^2 term does not disappear.

What happened with the ?
Use the quadratic formula.

Quote:

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Originally Posted by Siron

What happened with the ?
Use the quadratic formula.

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Not sure if this works?

- b + or - square root b^2 - 4ac / 2a

b = 4, a = -1, and c = 5

4^2 - 4 x 1 x 5 = 16 - 20 = - 4

square root of - 4 = E?

Quote:

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Originally Posted by David Green

Not sure if this works?

- b + or - square root b^2 - 4ac / 2a

b = 4, a = -1, and c = 5

4^2 - 4 x 1 x 5 = 16 - 20 = - 4

square root of - 4 = E?

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The discriminant is , so there are two solutions.

Of course, the easier method is to look at my post above and realise that the quadratic easily factorises.

Quote:

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Originally Posted by Prove It

The discriminant is , so there are two solutions.

Of course, the easier method is to look at my post above and realise that the quadratic easily factorises.

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I do keep getting my minus and a minus confused?

Thanks

In the second degree equation:

is that means if we substitute this values in the formula we get:

Quote:

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Originally Posted by Siron

In the second degree equation:

is that means if we substitute this values in the formula we get:

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Interestingly I still get confused with my rules?

36 squared = 6 / 2 = 3, 3 + (-4) = -1 first root

36 squared = 6 / 2 = 3, 3 - ( - 4) = 7 ?

How do I get the root 5?

There're two solutions (if ) for a quadratic equation by using the quadratic formula:

So in this case:

Quote:

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Originally Posted by Siron

There're two solutions (if ) for a quadratic equation by using the quadratic formula:

So in this case:

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I can't say that I can follow the reasoning there yet, what happend to the rest of the formula?

What do you mean with the 'rest of the formula'? Don't you understand how I got the solutions? ...

I understand the two roots and how you got them, but the formula has been modified.

- b + or - square root D / 2a

See the changes I have not seen before?

Capital D not squared, and - 4ac missing?

Like I said before this subject is new to me and is a new learning curve.

I don't think the formula has been modified.
In general if you have a quadratic equation:

Then this equation has:
two different solutions (or roots) if



If then the two identical solutions are:


If :
There're no real solutions.

Is this clear? If you need a prove of this then you can ask it.

So the capital D, this I now understand to be the solution of;

b^2 - 4ac

then the answer = D which I take the square root value.

D is the discriminant, and indeed .

I think you get the point now. Do you?