For which values of the number a (identity)?

For which values of the number a 

 xa + 2a = x + 2a^2 is an identity

The answer is going to be a=1. How can I get that answer? Can someone show me step by step because I want to learn. Thank you.

5 Answers

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  • ?
    Lv 7
    2 months ago

    ax+2a=x+2a^2

    =>

    x(a-1)=2a(a-1)

    =>

    (x-2a)(a-1)=0

    So, if a=1, then

    (x-2)*0=0

    =>

    0=0 is an identity, no matter

    what x is.

  • Ray S
    Lv 7
    2 months ago

    xa + 2a = x + 2a²                   ← We're asked to determine the value of "a".

                                                       So, we want to solve for "a", not x

              0 = x + 2a² - xa - 2a      ← Got everything on one side and equal to 0.

                                                        Now, begin factoring.

              0 = x - xa + 2a² - 2a       ← Rearranged the terms

              0 = x(1 - a) + 2a(a - 1)    ← Notice that (1-a) and (a-1)are opposites ...

                                                          i.e. -(1-a)=(a-1) and -(a-1)=(1-a)

                                                          So, +2a(a - 1) = -2a(1-a)

              0 = x(1 - a) - 2a(1 - a)     ← Now, factor out (1 - a)

              0 = (x - 2a) (1 - a )          ← Now, apply the Zero Product Property to solve for "a".

                       n/a   , 1- a = 0              Since x can be any number, we can't get a definite

    ANSWER →           a = 1              value for "a" by considering factor (x-2a).

                                                         But, from the factor (1-a), we get that a=1

                       Checking

                 xa + 2a = x + 2a²

              x*1 + 2*1 = x + 2*1²

                     x + 2 = x + 2     ✔

  • Pope
    Lv 7
    2 months ago

    xa + 2a = x + 2a²

    2a² + (-x - 2)a + x = 0

    This is a quadratic equation in a. Apply the quadratic formula.

    a = {x + 2 ± √[(-x - 2)² - 8x]} / 4

    a = {x + 2 ± √[(x - 2)²]} / 4

    a = [x + 2 ± (x - 2)] / 4

    a = x/2 or a = 1

    That first solution, a = x/2, is at least a little shaky, because a would not be a constant. Could we really call that a value of a when x itself is unknown?

    The solution a = 1 satisfies the condition without a doubt.

  • 2 months ago

    Equating coefficients of x we have:

    xa = x

    so, a = 1

    Also, equating constant terms we have:

    2a = 2a²

    Again, a = 1

    Then, with a = 1 we have:

    x + 2 ≡ x + 2

    :)>

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  • 2 months ago

    xa + 2a = x + 2a²

    xa - x = 2a² - 2a

    x.(a - 1) = 2a² - 2a

    x = (2a² - 2a)/(a - 1)

    x = 2a.(a - 1)/(a - 1)

    x = 2a

    a = x/2

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