Ask Question
26 February, 03:29

Write x^2 - 8x + 13 = 0 in the form (x - a) ^2 = b, where a and b are integers.

(x - 4) ^2 = 3

(x - 3) ^2 = 2

(x - 2) ^2 = 1

(x - 1) ^2 = 4

Can anyone show the steps on how to get the answer?

+4
Answers (1)
  1. 26 February, 05:49
    0
    Sure.

    The answer is (x-4) ^2 = 3

    Procedure

    Complete a perfect square trinomial to factor:

    x^2 - 8x + 13 = 0

    Notice that - 8x = 2 (x) (4)

    Given tha x is the square root of x^2, we want to include the square of 4, which is 16.

    We must just add 16 to the first two terms, which, of course, obliges to substract the same number to maintain the equality. This way:

    (x^2 - 8x + 16) + - 16 + 13 = 0

    Now the three terms in parenthesis can be factor as a binomial squared:

    (x - 4) ^2 + 13 - 16 = 0

    (x - 4) ^2 - 3 = 0

    (x - 4) ^2 = 3 Which is the answer.

    Let me know whether you could follow the explanation.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Write x^2 - 8x + 13 = 0 in the form (x - a) ^2 = b, where a and b are integers. (x - 4) ^2 = 3 (x - 3) ^2 = 2 (x - 2) ^2 = 1 (x - 1) ^2 = 4 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers