Question

Please help!!! Algebra 2 question

Answer 1

Hello from MrBillDoesMath!

 

Answer:

   4 + i

Discussion:

    Let's verify the square root given above:

   (4 + i) ^ 2= 4^2 + i^2 + 8i = 16 - 1 + 81 = 15 +8i

   so (4+i) is indeed a square root.

    How did I find this? Suppose that

    sqrt(15 + 8i) = a + bi

    Squaring both sides gives

       15 + 8i = a^2 + 2abi + b^2*i^2 or

       15 + 8i = a^2 + 2abi + b^2 (-1) or

       15 + 8i = (a^2 - b^2) + (2ab)i.

   Equating real and imaginary parts gives:

         a^2 - b^2 = 15 and    2ab = 8

  Now 2ab= 8 =>  b = 8/(2a) = 4/a. Substitute this value in the equation

  a^2 - b^2 = 15 to get

  a^2 - (4/a)^2 = 15.            Multiply both sides by a^2

  (a^2) * (a^2) - (4/a)^2 * (a^2) = 15 * a^2  or

  (a^4) - 16 =  15 a^2 or rearranging

  (a^2) ^2 - 15 (a^2) - 16 = 0.

  This is a quadratic equation in a^2! Use the quadratic formula to solve for

  a^2 . You will get that a^2  = -1 ( no way!  ) or a^2 = 16. If a^2 = 16, then

  a = 4 and from 2ab = 8, 2(4)b = 8, or 8b = 8 or b = 1.

I am now officially tired...... hope this helps.

Thank you,

MrB