Solve using the quadratic formula:?

Question

x2 – 3x + 3 = 5 this is the problem thanks for your help

Answer 1

For a quadratic equation in the form:

ax^2 + bx + c = 0

the quadratic formula is:
x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

step 1. put your equation in the general form (subtract 5 from both sides of the equation)

step 2. evaluate the quadratic formula using the coefficients a, b, c from your equation (1, -3, -2, respectively)

I expect to get [3 +/- sqrt(17)]/2

The problem with "I <3 AUG" answer is twofold:
the problem does not factor as she present it:
(x-2)(x-1)=x^2 - 3x + 2, whereas the problem has a -2

secondly, the problem wants you to use the quadratic formula, even though you can solve it with other methods. it is an exercise. you know: learn by doing.

Answer 2

question selection a million : For this equation x^2 - 7*x - a million = - 7 , answer right here questions : A. locate the roots utilising Quadratic formula ! B. Use factorization to hit upon the inspiration of the equation ! C. Use ending up the sq. to hit upon the inspiration of the equation ! answer selection a million : First, we ought to coach equation : x^2 - 7*x - a million = - 7 , right into a*x^2+b*x+c=0 type. x^2 - 7*x - a million = - 7 , circulate each and everything interior the splendid hand side, to the left hand fringe of the equation <=> x^2 - 7*x - a million - ( - 7 ) = 0 , it incredibly is the comparable with <=> x^2 - 7*x - a million + ( 7 ) =0 , now open the bracket and we get <=> x^2 - 7*x + 6 = 0 The equation x^2 - 7*x + 6 = 0 is already in a*x^2+b*x+c=0 type. In that type, we are able to actual derive that the fee of a = a million, b = -7, c = 6. 1A. locate the roots utilising Quadratic formula ! Use the formula, x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a) We had know that a = a million, b = -7 and c = 6, we ought to subtitute a,b,c interior the abc formula, with thos values. Which produce x1 = (-(-7) + sqrt( (-7)^2 - 4 * (a million)*(6)))/(2*a million) and x2 = (-(-7) - sqrt( (-7)^2 - 4 * (a million)*(6)))/(2*a million) it incredibly is the comparable with x1 = ( 7 + sqrt( 40 9-24))/(2) and x2 = ( 7 - sqrt( 40 9-24))/(2) Which make x1 = ( 7 + sqrt( 25))/(2) and x2 = ( 7 - sqrt( 25))/(2) So we get x1 = ( 7 + 5 )/(2) and x2 = ( 7 - 5 )/(2) So we've the solutions x1 = 6 and x2 = a million 1B. Use factorization to hit upon the inspiration of the equation ! x^2 - 7*x + 6 = 0 ( x - 6 ) * ( x - a million ) = 0 The solutions are x1 = 6 and x2 = a million 1C. Use ending up the sq. to hit upon the inspiration of the equation ! x^2 - 7*x + 6 = 0 ,divide the two side with a million Then we get x^2 - 7*x + 6 = 0 , all of us know that the coefficient of x is -7 we ought to apply the reality that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -7/2 = -3.5 So we've make the equation into x^2 - 7*x + 12.25 - 6.25 = 0 which could be became into ( x - 3.5 )^2 - 6.25 = 0 So we are able to get (( x - 3.5 ) - 2.5 ) * (( x - 3.5 ) + 2.5 ) = 0 by capacity of utilising the associative regulation we get ( x - 3.5 - 2.5 ) * ( x - 3.5 + 2.5 ) = 0 And it incredibly is the comparable with ( x - 6 ) * ( x - a million ) = 0 So we've been given the solutions as x1 = 6 and x2 = a million

Answer 3

Put everything on the left side:

x^2 - 3x - 2 = 0

This means a=1, b=-3, c=-2

Plug these into the quadratic formula

x = (-b +/- √(b^2-4ac))/2a

x = (3 +/- √(9+8))/2

x = (3 +/- √(17))/2

x = 3/2 + √(17)/2 , 3/2 - √(17)/2

Answer 4

We have a trinomial.

There are three terms:

term 1 is x^2...the number in front of x is 1.

term 2 is -3x...the number in front of x is -3

NOTE: to find term three, you must bring -5 over and add it to 3.
If you do this, you get -2 as term 3.

Our third term is -2.

We have:

a = 1

b = -3

c = -2

Plug the values for a, b and c into the quadratic formula and simplify.

For a more detailed explanation of the quadratic formula, go to:

http://www.purplemath.com/modules/quadform.htm

Guido

Answer 5

x^2-3x+3=5
x^2-3x-2=0

In this case we could already factor it instead of using the quadratic formula. We will only use the quadratic formula if the equation could not be factored easily.

x^2-3x-2=0
(x-2)(x-1)=0
x-2=0 or x-1=0
x=2 or x=1

The possible values of x would be 2 and 1.

Answer 6

you can use a calculator or
x^2 - 3x + 3 = 5 <=> x^2 - 3x -2 = 0
a= 1 ; b = -3 ; = c = -2
assume that delta = b^2 - 2a*c
x1 = ( -b + square root of delta ) / 2
x2 = ( -b - square root of delta ) /2

however there are some special cases :
if a+b+c = 0 then x1 = 1 ; x2 = c / a
if a - b +c =0 then x1 = -1 ; x2 = -c/a

Answer 7

the quadratic formula is as follows

-b (+-) square root of (b^2 - 2(a)(c))
divided by
2(a)

the number on front of x^2 is A
the number on front of x is B
the last number is C

so in this case a=1 b=-3 c=-2 (because you set everything to zero)

i hope that helped, i know the way i answered is probably a little confusing >.<

Answer 8

yeah thats what id do, but im pretty sure you'd need to change it to equal 0, x2-3x-2=0

Answer 9

(3+ sq rt. 17)/ 2 and (3 - sq rt 17)/ 2

x2 -3x-2=0
(-b +/- sq rt (b2 -4ac))/2a

Answer 10

do your own homework.

Hint: Can it fit in quadratic form?