• jxk@sh.itjust.works
    7·
    2 days ago

    If I’m correct, the answer is -25.

    They chose 169 because 169 × 4 is 26².

    (I got the 169 joke; I’m just solving this because I like to solve equations)

  • echolalia@lemmy.mlEnglish
    11·
    2 days ago

    (edit: I typed this up not as a response to OP, but for anyone who was curious about this but didn’t remember or never learned enough math to solve this for themselves. It had errors but I fixed them (I think, I just woke up and I’m tired of re-reading this comment). Sharing is imperative, and not everyone has had the chance to develop an appreciation for math).

    Here I come to explain the joke.

    The question asks for what values of b produce no real solutions.

    Quadratics (this is the type of equation this is) have two, one or no real real roots. (If we set x to some arbitrary number and the equation is equal to zero, we say x is a root of the equation. A real number is anything we traditionally call a number, like 2, 1/4, -248957.666667 and pi).

    So we need to find where it has only one real solution. Quadratics of the form (x-k)^2 = 0 have only one real solution, that solution is k.

    So, since 169 = 13^2 we can easily find the b that produces this (this part isn’t necessary as part of the solution, I include it in case it helps someone understand):

    -1(x-13)^2 = 0
-1(x^2 -26x + 169) = 0
-x^2 +26x - 169 = 0

    If we apply the quadratic formula, we can find what values of b produce imaginary roots. I’m not going to type the whole thing out, we just need to know what is happening under the “square root” part of the quadratic formula, the radicand:

    sqrt(b^2 - 4 * -1 * 169) 
sqrt(b^2 - 676)

    sqrt(676) is 26, by the way. So any value between 0 and 26 (or 0 and -26) will produce a negative value inside the radicand. We can’t take a square root of a negative number (unless you don’t mind that your solution has complex numbers), so any -26 < b < 26 will give us an equation without real solutions.

    Therefore, the least value of b is a value of b greater than -26, but less than 0. Since the question specifies it is an integer, that number is -25.

    • ChaoticNeutralCzech@feddit.orgEnglish
      3·
      2 days ago

      You neglected that the radicand features 𝑏² rather than 𝑏·|𝑏|, and the former rises as |𝑏| rises whether it’s positive or negative. Hence the equation having no real solutions for 𝑏²<676 implies |𝑏|<26, not 𝑏<26. As a result, 𝑏 ∈ (–26, 26) rather than 𝑏 ∈ (–∞, 26) satisfies the conditions, the smallest integer solution to the question is therefore 𝑏 = –25.

      • echolalia@lemmy.mlEnglish
        2·
        2 days ago

        I wrote my comment too hastily, you are correct. My comment has been edited, probably while you were typing yours.

        • ChaoticNeutralCzech@feddit.orgEnglish
          1·
          2 days ago

          You’ll need to edit it again. The solution range is 𝑏 ∈ (–26, 26), not 𝑏 ∈ (–26, 0) but yeah, the smallest integer is still –25.

          Also, the last paragraph is weirdly worded:

          the least value of b is a value of x greater than […]

          You don’t need to introduce some “𝑥” of yours (distinct from the quadratic equation’s 𝑥?!) to express –26 < 𝑏 < 26 in words.

    • Aatube@kbin.melroy.orgOP
      3·
      2 days ago

      b=-26 also produces roots, so the smallest possible value would be -25. I posted this because the question’s wording of “least possibility” confused me as I thought it meant probability.

  • RobotZap10000@feddit.nlEnglish
    9·
    2 days ago

    Woops, I just wrote out the entire solution, but forgot that sqrt(-676) has no real solutions indeed.

  • ChaoticNeutralCzech@feddit.orgEnglish
    3·
    2 days ago

    You could have edited the screenshot to contain 196 = 14² rather than 169 = 13² for more rule. The solution would then be –27 instead of –25 but the steps would be effectively the same.