Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

7x25x+13=07x^2-5x+13=0

Full solution

Q. 7x25x+13=07x^2-5x+13=0
  1. Calculate Discriminant: To determine the number of real solutions for the quadratic equation 7x25x+13=07x^2-5x+13=0, we can use the discriminant method. The discriminant (DD) is given by the formula D=b24acD = b^2 - 4ac, where aa, bb, and cc are the coefficients of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.
  2. Find Coefficients: For the given equation 7x25x+13=07x^2-5x+13=0, the coefficients are: a=7a = 7, b=5b = -5, and c=13c = 13. Let's calculate the discriminant (D)(D). \newlineD=(5)24(7)(13)D = (-5)^2 - 4(7)(13)
  3. Perform Calculations: Now, we perform the calculations:\newlineD=254(7)(13)D = 25 - 4(7)(13)\newlineD=25364D = 25 - 364
  4. Subtract Numbers: Subtracting 364364 from 2525 gives us:\newlineD=339D = -339
  5. Interpret Results: Since the discriminant DD is negative (D=339D = -339), this means that the quadratic equation 7x25x+13=07x^2-5x+13=0 has no real solutions. It has two complex solutions instead.

More problems from Multiply and divide rational numbers

QuestionGet tutor helpright-arrow

Posted 8 months ago