Quadratic formula help? No B and a question in general?

Question

I have a homework problem that has 49z(to the second power) = 1. I know you can bring the 1 over to C by subtracting it to both sides, but how do you solve B? Do you just do it without B? 

Also, I'm doing something wrong with the formula completely, apparently. 

For example I have for one problem:

x(to the second) + 5x - 6 = 0, so I wrote the formula as;

 -5 +/- (square root of) 5 (2nd) - 4*(1)*(-6) over 2.

I then got 5 (2nd) = 25, -4*1*-6 = 24.
Then adding those together; 49
49/2 = 24.5
Square root of 24.5 = 4.949
then:
-5 + 4.949 = -.051, and -5 - 4.949 = -9.949.

...according to my book, that isn't the answer >.<

Help please with either of these?

Aishmit · Accepted Answer

For the first question, you do not need b.

49z^2 -1 = 0

The easiest way to solve this is to use factoring.

49z^2 -1 = 0
(7z + 1)(7z -1) = 0
z = 1/7 or -1/7

The quadratic formula is:

-b +- √(b^2 - 4ac) / 2 = 0
Therefore,
= -5 +- √(25 +24) /2 
= -5 +- √(49) /2 
= -5 +- 7 /2

One is -5 + 7 /2 which is 
2/2 = 1

The other is -5 - 7 / 2 which is 
-12/2 = -6

Therefore the answer is x = 1 or -6

? · Answer

For   49z^2 = 1 -->  49z^2 - 1 = 0     then a = 49, b = 0, and c = -1

z = (-0 +- sqrt(0 - 4(49)(-1)))/2(49) = +- sqrt(196)/98  = +- 14/98 = +- 1/7


The second equation:  x^2 + 5x - 6 = 0 -->  a = 1, b = 5, and c = -6

x = (-5 +- sqrt(25 - 4(1)(-6)))/2 = (-5 +- sqrt(49))/2 = (-5 +- 7)/2  -->  (-5 + 7)/2 = 1  or (-5 - 7)/2 = -6

Makoto · Answer

for the first one if there is not B term that means B = 0 
so 4z^2 -1 = 0 
z = (-(0) +- sqrt(0^2 - 4(4)(-1))/2(4) 
z = sqrt(16)/8 
z = 4/8 
z = 1/2 

for the second one
x^2 +5x -6 = 0 
x = (-(5)+-sqrt((5)^2-4(1)(-6))/2(1)
x = (-5 +- sqrt(25 + 24) )/2 
x = (-5 +- sqrt(49))/2
x = (-5 +- 7 ) /2 
x = 1 OR x = -6

Quadratic formula help? No B and a question in general?

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