x^2-6x-41=0

Question

x^2-6x-41=0

 

Answer 1

Since 41 is a prime number, we can solve using (say) completing the square:

 

(x -3)^2 will FOIL to give x^2 -6x +9, so we must add that 9 to the right side.

 

so (x -3)^2 - 41 = 9    Now we'll add 41 to both sides

 

so       (x -3)^2  =  50  Now take the square root of both sides

 

so           (x -3)  =  ±√50 = ± 5√2, because 50 = 25*2, and √25 = 5. Now add 3 to both sides

 

so                 x  =  3  ± 5√2 ANSWER 

Answer 2

When you are solving for x in a quadratic equation and factoring using FOIL is not possible, it is a good idea to use the quadratic formula

 

x = (-b ± √(b² - 4ac)) / 2

 

and the plus/minus sign indicates that there will be two x values.

 

a, b, and c are constants in a quadratic equation

 

ax² + bx + c = 0

 

 

In this case:

 

a = 1

b = -6

c = -41

 

Plug in these numbers into the formula to get your two x values.