Help with homework..please :]?

Question

Quadratic Formula Help
(8th grade stuff)

the problem is x^2 + 5x - 6 = 0


I have to use the formula -b +or- the square root of b^2 - 4(a)(c) all over 2(a).

but I keep getting the wrong answer...

Help me out (the answer is x=1 and -6)

thank you

Answer 1

-5 +- sr [5^2-4(1)(-6)
----------------------------
2(1)

-5 +- sr [25 + 24]
------------------------
2

-5 +- sr [49]
----------------
2

-5 +- 7
---------
2

2 -12
-- or ------
2 2


answers 1, -6



sry if it looks confusing, i tried my hardest to make it understandable

Answer 2

Equation x^2 +5x - 6 = 0

Using the Quadratic Equation we have
a=1
b=5
c=6

x = -5 +,- ((25)-4 (1)(-6))^1/2 / 2

x = -5 +,- ((25 + 24))^1/2 / 2

x= -5 +,- (49)1/2 //2

x= (-5 +7 )/2

x = (-5-) + (-7)/2

x= 2/2 = 1

x= -12/2 = -6

Answer 3

a is the 1 in front of x², b is the 5 in front of x, c is the -6, so

x = -5/2 ± √(5² - 4(1)(-6)) / 2
x = -5/2 ± √(25 + 24) / 2
x = -5/2 ± √49 / 2
x = -5/2 ± 7/2
x = -5/2 + 7/2 = 2/2 = 1 or
x = -5/2 - 7/2 = -12/2 = -6

Answer 4

Using that formula u get,
[(-5) +/- sqroot {5*5 - 4*1*(-6)}]/(2*1)
= [(-5) +/- sqroot{25+24}]/2
= [(-5) +/- sqroot 49]/2
= [(-5) +/- 7]/2
(i.e.) (-5+7)/2 and (-5-7)/2 which is nothing but 1 and -6.

You wud ve forgotten the minus sign with 6. So u wud ve got -2 and -4 as ans wch is wrong.

Happy ??? ;-)

Answer 5

The formula applies to ax^2+bx+c=0

comparing this to the above equation we get
a=1 b=5 and c=-6

so x=(-5+sqrt(5^2 - 4*1*(-6)))/(2*1) or (-5-sqrt(5^2-4*1*(-6)))/(2*1)

x= (-5+sqrt(25+24))/2 or (-5-sqrt(25+24))/2
x=(-5+sqrt49)/2 or (-5-sqrt49)/2
x=(-5+7)/2 or (-5-7)/2
x=2/2 or -12/2
x=1 or -6

Voila!!

Answer 6

equation is x^2 + 5x - 6 = 0
a=1
b=5
c=-6
Evaluate +-sqrt(b^2 - 4ac)
=+- sqrt(5^2 - 4(1)(-6))
=+- sqrt(25 + 24)
=+- sqrt49
=+-7
now got to the formula
(-b +- sqrt(b^2 - 4ac))/2a and substitute values
(-5 +-7)/2
thus x =( -5+7)/2 or (-5-7)/2
= 2/2 or -12/2
= 1 or -6

Answer 7

Solution (Use Middle Term Breaking):
x^2 + 6x - x - 6 = 0
x (x + 6) -1 (x+6) = 0
(x - 1) (x + 6 ) = 0
Therefore,
( x - 1) = 0 OR ( x + 6 ) = 0
x = 1 OR x = - 6 Answer.

Answer 8

you could just factor

(x+6) (x-1)

x = -b +- Sroot(b^2 - 4ac)/ 2a

-5 +- sqrt(25+24)/ 2
-5+- 7 /2

2 /2 = 1

-12/2 = -6

Answer 9

x^2 + 5x -6 =0
<=> x^2+ 6x -1x -6=0
<=> (x^2 -1x) + ( 6x -6)=0
<=> x(x - 1) + 6(x-1)=0
<=> (x-1)(x+6) =0

<=> x-1 =0 or x+6=0
<=> x= 0+1 or x=0-6
<=> x=1 or x=-6
Then S={1;-6}

Answer 10

x=(-b +/- sqrt(b^2 -4ac))/2a... got that right.

x= (-5 + sqrt(5^2 -4(1)(-6))) / 2(1)
= (-5 + sqrt(25 +24))/2
= (-5 + sqrt(49))/2
= (-5 +7)/2
= (2)/2
=1

Going back,

x = (-5 -7)/2
= (-12)/2
= -6

Hope you see your mistake somewhere in there.

Answer 11

x=[-5+-(25 - (-24))^.5)]/2 remember 25-(-24)=49

x=(-5+7)/2 and x=(-5-7)/2

x=1 and x=-6