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Hi, Everyone: I have a job interview tomorrow where I must give a 10-min presentation on the quadratic formula for an intro class , where we are assuming students know both how to factor and how to complete the square. Please comment:
O.K. We are given an equation ax2+bx+c=0 , where a,b,c are real numbers, and we want to find the values of x that satisfy the equation -- if these values exist *, i.e., we want to find values of x , so that when we plug them in, in the formula, we get a 0. As an example, if we have x2+2x+1 , we sub-in , say, x=1 , and get:
12+2.1 +1= 4 ≠ 0 , so 1 does not work. We can see also that values like x=0 or x=1 don't work either.
We will see there is a formula that helps us find these values, and we will derive the formula. We will then see some examples of how to use it
* We're not allowing complex numbers yet.
The formula we will use is: -b/2a ±[[itex]\sqrt{b2-4ac}[/itex]]/2a
Let's see how we get the formula:
We start with the initial equation:
ax2+bx+c=0 , and assume a≠ 0 .
We then divide thru by a , to get:
x2+(b/a)x+c/a=0
We then go on to complete the square , by adding and subtracting (b/2a)2
to get:
x2+(b/a)x+ (b/2a)2- (b/2a)2 +c/a=0 , so:
(x+(b/2a))2= (b2/4a2)-c/a=
[b2-4ac]/4a2.
Now we solve for x, by first taking the square root:
x=-b/2a±[itex]\sqrt{b2-4ac}[/itex]/2a
And this does it.
Let's see some examples:
First , let's apply it to our original formula x2-2x+1:
here we have :
a=1 , b=-2 , c=1 , so we sub-in the formula to get:
x= -(-2)/2(1) ±√ [ (-2)2-4(1)(1)]/2(1)
=1± √0/2 , so we get that x is a double root .
Thanks for any comments.
Homework Statement
O.K. We are given an equation ax2+bx+c=0 , where a,b,c are real numbers, and we want to find the values of x that satisfy the equation -- if these values exist *, i.e., we want to find values of x , so that when we plug them in, in the formula, we get a 0. As an example, if we have x2+2x+1 , we sub-in , say, x=1 , and get:
12+2.1 +1= 4 ≠ 0 , so 1 does not work. We can see also that values like x=0 or x=1 don't work either.
We will see there is a formula that helps us find these values, and we will derive the formula. We will then see some examples of how to use it
* We're not allowing complex numbers yet.
Homework Equations
The formula we will use is: -b/2a ±[[itex]\sqrt{b2-4ac}[/itex]]/2a
The Attempt at a Solution
Let's see how we get the formula:
We start with the initial equation:
ax2+bx+c=0 , and assume a≠ 0 .
We then divide thru by a , to get:
x2+(b/a)x+c/a=0
We then go on to complete the square , by adding and subtracting (b/2a)2
to get:
x2+(b/a)x+ (b/2a)2- (b/2a)2 +c/a=0 , so:
(x+(b/2a))2= (b2/4a2)-c/a=
[b2-4ac]/4a2.
Now we solve for x, by first taking the square root:
x=-b/2a±[itex]\sqrt{b2-4ac}[/itex]/2a
And this does it.
Let's see some examples:
First , let's apply it to our original formula x2-2x+1:
here we have :
a=1 , b=-2 , c=1 , so we sub-in the formula to get:
x= -(-2)/2(1) ±√ [ (-2)2-4(1)(1)]/2(1)
=1± √0/2 , so we get that x is a double root .
Thanks for any comments.