the problem reads:

evaluate sqrt (-4) if possible.

my answer:

sqrt (-4)<0

therefore this is not a real number

looks correct

you cannot take the square root of a negative number in the real number set.

Have you learned the imaginary number i = sqrt(-1) ??

if so you can do:
sqrt(-4) = sqrt(4) x sqrt(-1)
= 2i

if you have not studied i, then you would say: "there is no real solution to sqrt(-4)"

  1. 👍 0
  2. 👎 0
  3. 👁 156
  1. 1/2(y-10)+y distributive propertys

    1. 👍 0
    2. 👎 0
    posted by jake

Respond to this Question

First Name

Your Response

Similar Questions

  1. Algebra

    Evaluate sqrt7x (sqrt x-7 sqrt7) Show your work. sqrt(7)*sqrt(x)-sqrt(7)*7*sqrt(7) sqrt(7*x)-7*sqrt(7*7) sqrt(7x)-7*sqrt(7^2) x*sqrt 7x-49*x ^^^ would this be my final answer?

    asked by Alexa on February 17, 2016
  2. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2,

    asked by COFFEE on July 10, 2007
  3. Math:)

    A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates

    asked by girly girl on March 22, 2018
  4. Calculus - Second Order Differential Equations

    Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8))

    asked by COFFEE on July 10, 2007
  5. some algebra help (radicals)

    I hope I am writing this down right.. I am trying to do some practice questions to learn 10^5 (sqrt)2y - 4^5 (sqrt)2y I am trying to figure out how to solve this They gave us some answers to choose from, but I am clueless on how

    asked by Cassie on May 6, 2007
  6. Math/Calculus

    Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks. y''+4y'+6y=0, y(0)=2, y'(0)=4 r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)(6))/2 r=-2 +/- sqrt(2)*i , alpha

    asked by COFFEE on July 12, 2007
  7. Math Help please!!

    Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine

    asked by Reggie on May 20, 2007
  8. Algebra 2: Radicals URGENT!!

    Could some kind, saintly soul help me solve this problem? Simplify: 8w sqrt(48w^5) - x^2 sqrt(3xw^2) . . =8w(√16)(√3)(√w^4)(√w) - x^2(√3)(√x)(√w^2) =32w^3(√3w) - wx^2(√3x) not much of a "simplification" really 8w

    asked by Maria on April 27, 2007
  9. math,algebra,help

    Directions are simplify by combining like terms. x radiacal 18 -3 radical 8x^2 can someone show me how to do these types of problems. thanks I cant determine the second term. For the first, I think you meant x sqrt(18) which

    asked by Jasmine20 on January 26, 2007
  10. Inequality

    When I solve the inquality 2x^2 - 6 < 0, I get x < + or - sqrt(3) So how do I write the solution? Is it (+sqrt(3),-sqrt(3)) or (-infinity, sqrt(3))? Why? Thanks. So would this work? abs x < ( sqrt 3 ) or - sqrt 3

    asked by Jen on November 5, 2006

More Similar Questions