For the equation x^2+6x=1 find exact roots

Question

For the equation x^2+6x=1 find exact roots

 

Answer 1

Hey Melissa -- we can re-frame x^2 +6x -1 ... (x^2 +6x +9) -10 ... (x+3)^2 -10 ...

 

x+3 = +/- sqrt 10 ==> x= -3 +/- sqrt 10 ... Best wishes, ma'am :)

Answer 2

Hi Melissa;

x²+6x=1

Let's bring positive 1 to the left and render the right side zero.

x²+6x-1=0

The quadratic equation is...

x=[-b+/- √(b²-4ac)]

            2a

 

The +/- is the best I could do for explaining plus-or-minus.

In your equation,

a, the coefficient of x², is 1.

b, the coefficient of x, is 6

c, the constant, is -1

 

x={-6+/- √[(6²-4(1)(-1)]}
               2(1)

 

x=-6+/- √(36+4)

           2

x=-6+/- √40

         2

Obviously, 40 cannot be square-rooted into an integer.  However, we can break this down into two numbers, 4 x 10.  The 4 can be square-rooted into the integer of 2.  Because the denominator, also known as the divisor, is 2, and because the 6 can be divided by 2, this would be best.

x=-6+/- √[(4)(10)]

             2

x=-6+/- 2√10

          2

x=-3+/- √10