please solve the equations: x^2 -6x =16?

8 and -2
1
.618 and -1.618

1) x^2 - 6x = 16 => x^2 - 6x - 16 = 0 => (x-8) (x+2) = 0 so
i) x-8=0=> x=8 and ii) x+2=0 => x=-2

2) 2x^2 -3x +1 = 0=> 2x^2 -2x -x + 1 =0=> 2x(x-1) -1(x-1) => (2x-1)(x-1) = 0 so
i) 2x-1=0=> x=1/2 ii)x-1=0=> x=1


3) x^2 -x -1 =0 a=1 b=-1 c=-1

x1,2 = (-b+-√D)/2a D= b^2 - 4ac = 1+4=5

=> x1,2 = (1+_ √5)/2 so x1 = (1+√5)/2 x2=(1-√5)/2

x^2 -6x - 16 = 0
x^2 - 8x + 2x - 16 = 0
( x - 8 ) ( x + 2) = 0
x=8 or x = - 2 ans.

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

x^2-x-1=0

x = ( ( 1 + ( 5) ^0.5) /2 = 1.618 or
x = ( ( 1 - ( 5) ^0.5) /2 = - 0.618 ans

x^2 - 6x - 16 = 0
(x - 8) * (x + 2)
x = -2 , 8


2x^2 - 3x + 1 = 0
(2x - 1) * (x - 1) = 0
x = 1/2 , 1


x^2 - x - 1 = 0

x = (1 +/- sqrt(5)) / 2

Question Number 1 :
For this equation x^2 - 6*x = 16 , answer the following questions :
A. Find the roots using Quadratic Formula !

Answer Number 1 :
First, we have to turn equation : x^2 - 6*x = 16 , into a*x^2+b*x+c=0 form.
x^2 - 6*x = 16 , move everything in the right hand side, to the left hand side of the equation
<=> x^2 - 6*x - ( 16 ) = 0 , which is the same with
<=> x^2 - 6*x + ( - 16 ) =0 , now open the bracket and we get
<=> x^2 - 6*x - 16 = 0

The equation x^2 - 6*x - 16 = 0 is already in a*x^2+b*x+c=0 form.
In that form, we can easily derive that the value of a = 1, b = -6, c = -16.

1A. Find the roots using 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)
Since a = 1, b = -6 and c = -16,
we just need to subtitute the value of a,b and c in the abc formula.
So x1 = (-(-6) + sqrt( (-6)^2 - 4 * (1)*(-16)))/(2*1) and x2 = (-(-6) - sqrt( (-6)^2 - 4 * (1)*(-16)))/(2*1)
Which is the same with x1 = ( 6 + sqrt( 36+64))/(2) and x2 = ( 6 - sqrt( 36+64))/(2)
Which make x1 = ( 6 + sqrt( 100))/(2) and x2 = ( 6 - sqrt( 100))/(2)
We got x1 = ( 6 + 10 )/(2) and x2 = ( 6 - 10 )/(2)
We get following answers x1 = 8 and x2 = -2


Question Number 2 :
For this equation 2*x^2 - 3*x + 1 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !

Answer Number 2 :
The equation 2*x^2 - 3*x + 1 = 0 is already in a*x^2+b*x+c=0 form.
By matching the constant position, we can derive that the value of a = 2, b = -3, c = 1.

2A. Find the roots using Quadratic Formula !
By using abc formula the value of x is both
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
As a = 2, b = -3 and c = 1,
we just need to subtitute the value of a,b and c in the abc formula.
So we get x1 = (-(-3) + sqrt( (-3)^2 - 4 * (2)*(1)))/(2*2) and x2 = (-(-3) - sqrt( (-3)^2 - 4 * (2)*(1)))/(2*2)
Which is the same with x1 = ( 3 + sqrt( 9-8))/(4) and x2 = ( 3 - sqrt( 9-8))/(4)
Which is the same with x1 = ( 3 + sqrt( 1))/(4) and x2 = ( 3 - sqrt( 1))/(4)
We got x1 = ( 3 + 1 )/(4) and x2 = ( 3 - 1 )/(4)
We get following answers x1 = 1 and x2 = 0.5


Question Number 3 :
For this equation x^2 - x - 1 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !

Answer Number 3 :
The equation x^2 - x - 1 = 0 is already in a*x^2+b*x+c=0 form.
In that form, we can easily derive that the value of a = 1, b = -1, c = -1.

3A. Find the roots using Quadratic Formula !
By using abc formula the value of x is both
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
As a = 1, b = -1 and c = -1,
we just need to subtitute the value of a,b and c in the abc formula.
So x1 = (-(-1) + sqrt( (-1)^2 - 4 * (1)*(-1)))/(2*1) and x2 = (-(-1) - sqrt( (-1)^2 - 4 * (1)*(-1)))/(2*1)
Which make x1 = ( 1 + sqrt( 1+4))/(2) and x2 = ( 1 - sqrt( 1+4))/(2)
Which is the same as x1 = ( 1 + sqrt( 5))/(2) and x2 = ( 1 - sqrt( 5))/(2)
So we get x1 = ( 1 + 2.23606797749979 )/(2) and x2 = ( 1 - 2.23606797749979 )/(2)
The answers are x1 = 1.61803398874989 and x2 = -0.618033988749895

I'm 100% on the first 2, but you might want to double check the last one. I'm only about 75% on that.


x^2 - 6x = 16

subtract 16 from each side, so you can factor the polynomial

x^2 - 6x - 16 = 0

factor the polynomial

(x-8)(x+2) = 0

x = 8,-2

It's basic Algebra. You might want to practice it more.


next

2x^2 - 3x + 1 = 0

No need to get all the terms on one side, so you can just factor the polynomial.

(2x - 1)(x - 1) = 0

x = 1, 1/2

It's basic Algebra. You might want to practice it more.


final one

x^2 - x - 1 = 0

Again, the terms are already on one side, so you just need to factor the polynomial. However, since there are no factors of (-1) and (1) that will add up to equal (-1), there is no solution to this one.


Remember the quadratic formula.

To solve Ax^2 + Bx + C = 0

use this formula

(-B ± root (B^2 - 4AC)) / 2A

now use that to solve x^2 - x - 1 = 0

( -(-1) ± root ((-1)^2 - 4(1)(-1))) / 2(1)

(1 ± root (1 - (-4)) / 2

(1 ± root (5)) / 2

x = ((1 - root 5) / 2) or ((1 + root 5) / 2)

Okay, so that last one wasn't basic Algebra, but I still remembered how to do it 15 years out of high school.

1)
x^2 - 6x = 16
x^2 - 6x - 16 = 0
x^2 + 2x - 8x - 16 = 0
(x^2 + 2x) - (8x + 16) = 0
x(x + 2) - 8(x + 2) = 0
(x + 2)(x - 8) = 0

x + 2 = 0
x = -2

x - 8 = 0
x = 8

∴ x = -2, 8
_______________________

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

2x - 1 = 0
2x = 1
x = 1/2 (0.5)

x - 1 = 0
x = 1

∴ x = 1/2 (0.5), 1
_______________________

3)
x^2 - x - 1 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)

a = 1
b = -1
c = -1

x = [1 ±√(1 + 4)]/2
x = [1 ±√5]/2

∴ x = [1 ±√5]/2

Qu 1
x^2 - 6 x - 16 = 0
( x - 8 ) ( x + 2 ) = 0
x = 8 , x = - 2

Qu 2
2x^2 - 3x + 1 = 0
( 2x - 1 ) ( x - 1 ) = 0
x = 1/2 , x = 1

Qu 3
x^2 - x - 1 = 0
x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a
x = [ 1 ± √ ( 1 + 4 ) ] / 2
x = [ 1 ± √5 ] / 2

x^2-6x=16
x^2-6x-16=0
(x-8)(x+2)=0
x=8 x=-2

2x^2-3x+1=0
(2x-1)(x-1)=0
2x-1=0 x-1=0
x=1/2 x=1

x^2-x-1=0
use quadratic formula

x^2 - 6x = 16

==> x^2 -6x - 16 = 0

==> x^2 -8x + 2x - 16 = 0

==> x(x - 8) + 2(x - 8) = 0

==> (x + 2) (x - 8) = 0

==> x = -2 or x = 8

x^2 - 6x - 16 =0
(x - 8)(x + 2) = 0
x = 8 or -2


2x^2 - 3x + 1 = 0
(2x - 1)(x - 1) = 0
2x - 1 = 0 or x - 1 = 0
x = 1/2 or 1.

x^2 - x - 1 = 0
No easy factors. Use x = (-b +-sqrt(b^2 - 4ac))/2a


x = (-(-1) +-sqrt((-1)^2 - 4(1)(-1)))/2(1)


x = (1 +-sqrt(1 +-4)/2 = ( 1 + sqrt5)/2 or (1 - sqrt5)/2