Can you help me solve x^2-3x+11=0?

Question

thank you all

Answer 1

please note to the one who answered before me that 9 - 44 = -35 and not -36

*****

by quadratic formula,
a = 1, b = -3, c = 11

x = [-b +/- √(b^2 - 4ac) ] / 2a
= [ -(-3) +/- √((-3)^2 - 4*1*11) ] / 2*1
= [ 3 +/- √(9 - 44) ] / 2
= [ 3 +/- √(-35) ] / 2
= [ 3 +/- √(35) i ] / 2

Answer 2

x^2 - 3x + 11 = 0

Using the quadratic formula:

x = [3 +/- √(9 - 44)]/2
x = [3 +/- √(-35)]/2

There is a small problem here. Square root of a negative number is not defined. So, this quadratic equation has no real roots. I'll solve it further though, and give the imaginary answer in terms of i, where i = √(-1)
x = [3 +/- √(-35)]/2
x = [3 +/- i√35]/2
x = (3 + i√35)/2 and x = (3 - i√35)/2 are the roots of the equation.

Answer 3

Solve,

x² - 3x + 11 = 0

x² - 3x = -11

x² - 3x + (3/2)² = -11 + (3/2)²

(x - 3/2)² = -11 + 9/4

(x - 3/2)² = (-44 + 9)/4

(x - 3/2)² = (-35)/4

x - 3/2 = ± √(-35)/2

x = 3/2 ± √(-35)/2

x = [ 3 ± √(35)i ]/2

It's has complex roots.

Answer 4

x^2 - 3x + 11 = 0

We can solve this by using the quadratic formula. As a reminder, the quadratic formula goes as follows:

For an equation ax^2 + bx + c = 0,
x = [ -b +/- sqrt(b^2 - 4ac ] / (2a)

Plug-and-chug. a = 1, b = -3, c = 11, so

x = [ -(-3) +/- sqrt( (-3)^2 - 4(1)(11) ) ] / (2*1)
x = [ 3 +/- sqrt(9 - 44) ] / 2

From here, we can see that we will end up with the square root of a negative number, which means we will have no real solutions. We will, however, have complex solutions (two, actually).

x = [ 3 +/- sqrt(-36) ] / 2

x = [ 3 +/- sqrt( -1 * 36 ) ] / 2

x = [ 3 +/- sqrt(-1)sqrt(36) ] / 2

By definition, sqrt(-1) = i, so

x = [ 3 +/- i * 6 ] / 2

x = [3 +/- 6i ] / 2

x = (3/2) +/- (6i)/2

x = (3/2) +/- 3i

So the two solutions are

x = { (3/2) + 3i , (3/2) - 3i }

Answer 5

x ^2 - 3x +11= 0

by using quadratic formula

x = - b +/- square root b ^ 2 -4 a c divide by 2 a

x = -(-3) +/- square root (-3)^2-4(1)(11)
/ 2

= 3 +/- square root 9 - 44 / 2

= 3+/- square root -35 / 2

= 3 +/- 5.92 /2

either x = 3 + 5.92/2 = 4.46

or x= 3 - 5.92 = -1.46

Answer 6

try solvin it using quadratic method
ie compare the eqaution with ax^2 +bx+c...where the values of a is 1 ,b is -3, and cis 11,there fore the roots of the eqaution i.e (there will be two roots bcoz its eqaution of order 2)rt will be x= (-b+squarertof(b^2-4ac))/2a.......and x=(-b-squarert(b^2-4ac))/2a........substitute the values and ull get the answers!!

Answer 7

(Your question)-----------x^2 - 3x + 11 = 0
(Bring 11 over to the other side)--- x^2 - 3x = -11
(Complete the square)------- (x - 3/2)^2 - (3/2)^2 = -11
------------------------------- (x - 3/2)^2 - 9/4 = -11
(Bring -9/4 over to the other side)-- (x - 3/2)^2 = -11 + 9/4
------------------------------------- (x - 3/2)^2 = 7/4
(Square root (s.r.) both sides)--- x - 3/2 = (+or-) s.r.(7/4)
(Bring -3/2 over to the other side)-- x = 3/2 (+or-) s.r.(7/4)
------------------------------ x = [3/2 + s.r.(7/4)] or [3/2 - s.r.(7/4)]
Hence, x= 2.823 or x= 0.177

Answer 8

x^2 -3x +11 = 0
x^2 -3x = -11
x( x- 3) = -11
x = -11/ x-3

Answer 9

Use quadratic equation formula since the equation can not be factored.

Answer 10

You'd have to use the quadratic formula.
[-b +/- (root(b^2-4ac))/2a]