algebra

posted by .

solve for u, u=sqrt -u+6

u=ã-u+6

  • algebra -

    do you mean

    u = (sqrt(-6 + u))

    please use the word 'sqrt' and put in ()
    what is included under the radical

    we will be able to assist you much faster if what you need is more clear

    thank you

  • algebra -

    u = (sqrt(-u + 6))

    u = (sqrt(6 - u))
    square both sides
    u^2 = 6 - u
    u^2 + u - 6

    complete the square
    u^2 + u = 6
    u^2 + u + 1/4 = 6 + 1/4
    (u + 1/2)^2 = 25/4
    take the square root of both sides

    + - (u + 1/2) = 5/2
    u + 1/2 = 5/2
    u = 4/2 = 2

    -u - 1/2 = 5/2
    -u = 6/2 = 3
    u = -3

    check for u = 2
    u = (sqrt(6 - u))
    2 = (sqrt(6 - 2))
    2 = sqrt 4
    2 = 2

    check for u = -3
    u = (sqrt(6 - u))
    -3 = (sqrt(6 + 3))
    -3 = sqrt 9
    -3 not = 3

    so, the only solution is u = 2

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Algebra

    Solve for s: h=(square root of 3)times s/2 and solve for h V= (pi)r squared h / 3 Solve for s: h=(square root of 3)times s/2 Multiply both sides by 2. 2h = (sqrt 3)*s*2/2 which cancels the 2 on the right. 2h = (sqrt 3)*s Now divide …
  2. Inequality

    When I solve the inquality 2x^2 - 6 < 0, I get x < + or - sqrt(3) So how do I write the solution?
  3. math

    solve 2x^2+3x+8=0 and express the solutions in a+bi form. Let's use the quadratic formula to solve for x and express those solutions in a+bi form. x = [-b + or - sqrt(b^2 - 4ac)]/2a Note: sqrt = square root. a = 2, b = 3, and c = 8 …
  4. Algebra 2: Radicals URGENT!!

    Could some kind, saintly soul help me solve this problem?
  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 to …
  6. 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 …
  7. 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?
  8. algebra

    can someone tell me if i did the problem correct (solve by completing the square) 4x^2+2x-3=0 x^2+(1/2)=3/4 x^2+(1/2)+(1/4)^2=3/4+(1/4)^2 (x+1/4)^2=13/16 x+1/4=+-sqrt 13/16 x+1/4=+-sqrt 13/4 x=-1/4+-sqrt 13/4 x=-1+-sqrt 13/4
  9. Algebra 2

    Operations with Complex Numbers Simplify. 1. sqrt(-144) 2. sqrt(-64x^4) 3. sqrt(-13)*sqrt(-26) 4. (-2i)(-6i)(4i) 5. i13 6. i38 7. (5 – 2i) + (4 + 4i) 8. (3 – 4i) – (1 – 4i) 9. (3 + 4i)(3 – 4i) 10. (6 – 2i)(1 + i) 11. (4i)/(3+i) …
  10. 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?

More Similar Questions