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Solve 3x^2-8x+4=0 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat...Asked by Jon
Solve 3x^2-8x+4=0 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.
A)2;between 0 and 1
B)between 0 and 1;between 7 and 8
C)1,2
D)between 0 and 1;between 3 and 4
I don't understand this one either its gotta be A or C I don't undertand what I got from a website which gave me X1=2 and X2=.6 repeating
A)2;between 0 and 1
B)between 0 and 1;between 7 and 8
C)1,2
D)between 0 and 1;between 3 and 4
I don't understand this one either its gotta be A or C I don't undertand what I got from a website which gave me X1=2 and X2=.6 repeating
Answers
Answered by
Reiny
let y = 3x^2-8x+4 or f(x) = 3x^2-8x+4 , same thing
try a few x's
I tried -1,0,1,2,3 and y values of
15,4,-1,0,7
notice when x = 0 the graph is above the x-axis, but when x=1 the graph is below the x-axis.
So there has to be a solution between x=0 and x=1
Also notice when x=2 , y =0, so x=2 is a solution.
<b>Solutions to equations are where the matching graph cuts the x-axis.</b>
Now that we know that we have an exact solution at x=2, x-2 must be an exact factror, and sure enough
3x^2-8x+4 = (x-2)(3x-2)
so x=2 or x=2/3
try a few x's
I tried -1,0,1,2,3 and y values of
15,4,-1,0,7
notice when x = 0 the graph is above the x-axis, but when x=1 the graph is below the x-axis.
So there has to be a solution between x=0 and x=1
Also notice when x=2 , y =0, so x=2 is a solution.
<b>Solutions to equations are where the matching graph cuts the x-axis.</b>
Now that we know that we have an exact solution at x=2, x-2 must be an exact factror, and sure enough
3x^2-8x+4 = (x-2)(3x-2)
so x=2 or x=2/3
Answered by
Jon
So C would be the choice?
Answered by
Reiny
NOOO, it is A
Did you not read when I said:
"notice when x = 0 the graph is above the x-axis, but when x=1 the graph is below the x-axis.
So there has to be a solution between x=0 and x=1
Also notice when x=2 , y =0, so x=2 is a solution. "
Did you not read when I said:
"notice when x = 0 the graph is above the x-axis, but when x=1 the graph is below the x-axis.
So there has to be a solution between x=0 and x=1
Also notice when x=2 , y =0, so x=2 is a solution. "
Answered by
Jon
Oh yeah duh. my mistake
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