Math help!! urgent!?

Question

Solve the following equations using either the quadratic formula or by completing the square.Give your answers to 2 d.p.

x2-35=0

and also
x2-3x-3=0[am i supposed to use the square for this?how exactly do you do this?]

I know im prob sposed to work these out myself but they're during tmr and im really stuck!
its one of my GCSE practice questions.There are bout 20 more but I just need to know the basics in order to do the rest.
thanks a mill.

Luvv,
Kris.

Answer 1

1st question.
you can just simply do
x^2 = 35 or x = (+/-) sqrt(35) = (+/-)(5.92)

2nd question
I would just use the quadratic formula because it not an obvious polynomial that can simply be factored. Plus the quadratic formula can use used for any quadratic equation.
So the solutions are x = (1/2)*(3 +/- sqrt(21)) = 3.79, -0.79

Answer 2

1.X^2-35=0
X=+ or -(sqrt 35)

2.X^2-3X-3=0
Let a=1 b=-3 c=-3
Use: (-b +or- [sqrtb^2-4ac])divided by 2a
No need to worry about +or-
e.g how to solve +or-
At last, for example you've got 5x+or-3
So the answer is 2: 1.5x+3
2.5x-3

Answer 3

x^2-35=0
x^2=35
x= square root of 35 =5.916

determinant of x^2-3x-3 =0 is 9-(4*1*(-3))=21

roots are( 3+/-square root 21)/2

x1=-0.791 x2 =3.791

Answer 4

well sweetie , x^2-35=0
x^2=35
x= square root of 35 =5.916

determinant of x^2-3x-3 =0 is 9-(4*1*(-3))=21

roots are( 3+/-square root 21)/2

x1=-0.791 x2 =3.791

Good luck . good question . -B-

Answer 5

do you know what the quadratic equation is?

x=[-b+-sqrt(b^2-4ac)]/2a

a b and c are the coefficients in front of the variables. (in the case of x^2-3x-3, you have 1, -3, and -3.)
try plugging in the values for a b and c and it should work.

good luck

Answer 6

x^2-35=0
x^2=35
x=+-sqrt35 [square rooting both sides]
=+-5.92

x^2-3x-3=0
x^2-3x=3
(x)^2-2*x*3/2+(3/2)^2=3+9/4 [adding (3/2)^2 or 9/4 to both sides]
(x-3/2)^2=21/4
x-3/2=+-sqrt21/2 [square-rooting both sides]
x=3/2+-2.29
=1.5+2.29 or 1.5-2.29
=3.79 or -0.79