Solving quadratic equations using the quadratic formula? Please help!?
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Solving quadratic equations using the quadratic formula? Please help!?

[From: ] [author: ] [Date: 17-04-11] [Hit: ]
https://www.youtube.com/watch?v=E5L-LPXS.........

You might find this video helpful.
https://www.youtube.com/watch?v=E5L-LPXS...

Bear in mind though just because there is a value greater than one in front of the x squared term doesn't mean you can't factorise it.

Look for two numbers that multiply to give 7*-7 = -49 and add together to give -48, clearly that's -49 and 1. Now decompose your x terms into -49x + x. so our first equation becomes:

7v^2 - 49v + v -7 = 0
now factor the first and last two terms.

7v(v-7) + 1(v-7) = 0
now factor the common term.
(v-7)(7v+1) = 0
so either v-7 = 0 or 7v+1 = 0
in which case v = 7 or v = -1/7

You might also find this video helpful.

https://www.youtube.com/watch?v=XNuEtbiZ...
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RAT M say: 22
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Raymond say: You have two choices.

The easy one is to use the full quadratic formula, which takes care of the "leading coefficient" (the number in front of the highest-degree term).

The equations already fit the standard format of:

av^2 + bv + c = 0
where (according to the quadratic formula),
v = [ -b ±√( b^2 - 4ac )] / 2a

In the first equation, you have
a = 7
b = -48
c = -7

You just use these numbers, for a, b and c, in the formula.
Careful with signs (for example -b = -(-48) = +48)

The "a" takes care of the "number in front".

v = [ -b ±√( b^2 - 4ac )] / 2a
becomes
v = [ +48 ±√( 2304 + 196 )] / 14
v = [ 48 ± 25 ] / 14

Using +
v = (48+25)/14 = 73/14 = 5 and 3/14
and so on.

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The second choice you have is to divide the entire equation (every term, both sides) by the leading coefficient. This new equation will be "monic" (its leading coefficient will be 1) AND it will have the same solutions - the same values for the variable - than the original equation.

For example, in the third one:

25x^2 + 10x + 1 = 0
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