Solve for 'x'?

Question

x^2=5x+2

Answer 1

x^2-5x-2=0, by quadratic equation, x= 5-sqrt(33)/2,5+sqrt(33)/2

Answer 2

Equivalently, this is:

x^2 - 5x - 2 = 0

In other words, you want to find the "roots" to the quadratic polynomial x^2 - 5x - 2. You can use the quadratic formula for this. The quadratic formula says that for any quadratic:

a*x^2 + b*x + c

the formulas for its two roots (assuming a is not 0) are:

( -b + sqrt( b^2 - 4*a*c ) )/ (2 * a)

and

( -b - sqrt( b^2 - 4*a*c ) )/ (2 * a)

where sqrt(y) is the square root of y. In your case, a=1, b=-5, and c=-2. So, your two roots are:

( 5 + sqrt( 25 - 4*(-2) ) )/2 = ( 5 + sqrt(25+8) )/2 = ( 5 + sqrt(33) )/2

and:

( 5 - sqrt(33) )/2

In other words, both:

x = ( 5 - sqrt(33) )/2

and

x = ( 5 + sqrt(33) )/2

will solve your problem. To verify, for x = ( 5 + sqrt(33) )/2,

( ( 5 + sqrt(33) )/2 )^2 = 25/4 + 10*sqrt(3)/4 + 33/4 = 58/4 + 5*sqrt(3)/2 = 29/2 + 5*sqrt(3)/2

5*( 5 + sqrt(33) )/2 + 2 = 25/2 + 5*sqrt(33)/2 + 2 = 29/2 + 5*sqrt(33)/2

and for x = ( 5 - sqrt(33) )/2,

( ( 5 - sqrt(33) )/2 )^2 = 25/4 - 10*sqrt(3)/4 + 33/4 = 58/4 - 5*sqrt(3)/2 = 29/2 - 5*sqrt(3)/2

5*( 5 - sqrt(33) )/2 + 2 = 25/2 - 5*sqrt(33)/2 + 2 = 29/2 - 5*sqrt(33)/2

So in both cases, x^2 = 5x + 2.

Answer 3

x^2-5x-2=0 , D=33, x=5+33/2 or x=5-33/2 (33 squared rute)

Answer 4

set it equal to 0: 0=x^2-5x-2
apply the quadratic formula:
(25-4*-2*1)^1/2=10*2^1/2
(5+or -10*2^1/2)/2=2.5-5*2^1/2 and 2.5+5*2^1/2

The answer are x=9.571067812 and x= -4.571067812

Answer 5

5+or- the squared root of 33 over 2

Answer 6

x1=2,5+sqrt(33)/2
x2=2,5-sqrt(33)/2

Answer 7

You need to do your own homework problems....

Answer 8

x= do it yourself