Question:

How to solve this? 3(x-5)^2 - 12 = 0?

by  |  earlier

0 LIKES UnLike

How to solve this? 3(x-5)^2 - 12 = 0?

 Tags:

   Report

4 ANSWERS


  1. 3(x - 5)^2 - 12 = 0

    3(x - 5)^2 = 12

    (x - 5)^2 = 12/3

    (x - 5)^2 = 4

    x - 5 = ±√4

    x - 5 = ±2

    x - 5 = 2

    x = 2 + 5

    x = 7

    x - 5 = -2

    x = -2 + 5

    x = 3

    ∴ x = 3 , 7


  2. 3(x^2-10x+25) - 12 = 0

    3x^2 - 30x + 75 - 12 = 0

    3x^2 -30x + 63 = 0

    x^2 - 10x + 21 = 0

    (x - 7)(x - 3) = 0

    x - 7 = 0   or x - 3 = 0

    x = 7 or x = 3

  3. 3 (x - 5)² = 12

    (x - 5)² = 4

    x - 5 = ± 2

    x = 7 , x = 3

  4. 3(x-5)² - 12 = 0

    => 3(x² + 5² - 2*5*x) - 12 = 0  [because (a-b)² = a² + b² - 2ab]

    => 3[(x² + 5² - 2*5*x) - 4] = 0 [taking 3 common from the whole equation]

    => (x² + 5² - 2*5*x) - 4 = 0/3 = 0

    => x² - 10x + 25 - 4 = 0

    => x² - 10x + 21 = 0

    => x² - 3x - 7x + 21 = 0 [splitting the middle term]

    => x(x-3) - 7(x-3) = 0

    => (x-3) (x-7) = 0

    x-3 = 0

    => x = 3

    OR

    x-7 = 0

    => x = 7

    so the two values of x, on solving the quadratic equation are x = 7 and x = 3

Question Stats

Latest activity: earlier.
This question has 4 answers.

BECOME A GUIDE

Share your knowledge and help people by answering questions.
Unanswered Questions