PLEASE HELP! Simple math question?

I’m confused with how to work out “a” in this equation..

a + root a =132

15 Answers

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  • 2 months ago

    (I'm assuming your use of "root" is the principal square root denoted by the √ symbol.)

    I solved it using a substitution:

    a + √a = 132

    √a(√a + 1) = 132

    Let x = √a

    x(x + 1) = 132

    At this point, we can solve by inspection knowing that 11 * 12 = 132 or -12 * -11 = 132

    That means x = 11 or x = -12

    Since x = √a → a = x²

    a = 11² = 121

    or

    a = (-12)² = 144

    However in testing the answers, we see that 144 is an extraneous result and reject it.

    121 + √121 =? 132

    121 + 11 = 132 ✓ (accept a = 121)

    144 + √144 =? 132

    144 + 12 =? 132

    156 ≠ 132 (reject a = 144)

    Answer:

    a = 121

    If you didn't notice the way to solve for x by inspection, you could solve it by factoring:

    x(x + 1) = 132

    x² + x - 132 = 0

    (x + 12)(x - 11) = 0

    Hence:

    x = -12 or x = 11

    a = 144 or a = 121

    Again reject 144 by testing.

    Answer:

    a = 121

    Update: The radical sign (√) is defined to return the principal (non-negative) square root. 

  • 2 months ago

    ==> sqrt(a) = 132 - a ... now square both sides to get rid of sqrt(a)

    ==> a = (132-a)^2

    ==> a = 132^2 - 264a + a^2

    ==> a^2 - 265a + 132^2 = 0

    solve for a

    ==> a = (265 +/- sqrt(265^2 - 4* 132^2))/ 2

    ==> a = (265 +/- sqrt(265^2 - 264^2))/2

    ==> a = (265 +/- sqrt((265 + 264)(265 -264)))/2

    ==> a = (265 +/- sqrt(529))/2

    ==> a = (265 +/- 23)/2

    ==> a = 242/2 or 188/2

    ==> a = 121 or 144

  • 2 months ago

    this picture should help:  

    Attachment image
  • 2 months ago

    a = 121 root a = 11

    It is just by looking at numbers bigger than 10.The first one,11,gives the answer.

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  • sepia
    Lv 7
    2 months ago

    a + root a = 132

    sqrt (a) = 132 - a

    a = (132 - a)^2

    a = (a - 132)^2

    Solutions:

    a = 121

    a = 144

  • Ian H
    Lv 7
    2 months ago

    Let a = x^2, (as suggested by 121 + 11 = 132)

    x^2 + x – 132 = 0 = (x – 11)(x + 12)

    a was 121

  • 2 months ago

    a + root a = 132

    sqrt (a) = 11

    a = 121

  • 2 months ago

    If that is:

    a + √a = 132

    Let's first subtract "a" from both sides to get the radical by itself:

    √a = 132 - a

    Now square both sides:

    a = (132 - a)²

    Expand the right side and simplify into a quadratic:

    a = 17424 - 264a + a²

    a = a² - 264a + 17424

    0 = a² - 265a + 17424

    Now use the quadratic equation to find possible solutions for a:

    a = [ -b ± √(b² - 4ac)] / (2a)

    a = [ -(-265) ± √((-265)² - 4(1)(17424))] / (2 * 1)

    a = [ 265 ± √(70225 - 69696)] / 2

    a = (265 ± √529) / 2

    a = (265 ± 23) / 2

    a = 242/2 and 288/2

    a = 121 and 144

    Now we test both possible solutions to see if any do not make sense:

    a + √a = 132

    121 + √121 = 132 and 144 + √144 = 132

    121 + 11 = 132 and 144 + 12 = 132

    132 = 132 and 156 = 132

    TRUE and FALSE

    Since the second is FALSE, we can throw that out as an extraneous solution.  There is one solution to your equation:

    a = 121

  • Let sqrt(a) = x.  Now you have:

    x^2 + x = 132

    It's just a quadratic

    x^2 + x - 132 = 0

    x = (-1 +/- sqrt(1 + 4 * 132)) / 2

    x = (-1 +/- sqrt(1 + 528)) / 2

    x = (-1 +/- sqrt(529)) / 2

    x = (-1 +/- 23) / 2

    x = -24/2 , 22/2

    x = -12 , 11

    sqrt(a) = -12 , 11

    a = (-12)^2 , 11^2

    a = 144 , 121

    Some will try to say that a = 144 is extraneous, because if you graph the function, non-principal roots aren't defined.  Of course, they fail to point out when or where this problem ever said that this was a function.

    a = 144 , 121

    Try them out.

  • Tasm
    Lv 6
    2 months ago

    I assume it is square root. 

    root a = 132 - a

    a = (132-a) ^2

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