Algebra

It is possible for a quadratic equation to have no real-number solutions:Solve : x2 + 5x + 3 = 0

  1. 👍 0
  2. 👎 0
  3. 👁 46
asked by tom
  1. Sure, that just means that the parabola never crosses the x axis but has a vertex above if it opens up or below if it opens down.

    x^2 + 5 x + 3

    x = [ -5 +/- sqrt (25 - 12) ] / 2

    that has real roots because (b^2-4ac) is positive.
    However try
    x^2 + 2 x + 3 = 0

    x = [ -2 +/- sqrt (4 - 12) ]/2

    x = -1 +/- .5 sqrt(-8)
    sqrt of -8 = i sqrt 8 = 2i sqrt(2)

    x = -1 +/- i sqrt 2
    the solutions are complex because they contain i, the sqrt of -1

    1. 👍 0
    2. 👎 0
    posted by Damon

Respond to this Question

First Name

Your Response

Similar Questions

  1. Algebra II

    1)What method(s) would you choose to solve the equation: x2 + 2x - 6 = 0 A. Square roots; there is no x-term. B. Quadratic formula, graphing; the equation cannot be factored easily since the numbers are large. C. Factoring; the

    asked by Hally on January 15, 2013
  2. algerba

    What is the value of the discriminant, b2 − 4ac, for the quadratic equation 0 = −2x2 − 3x + 8, and what does it mean about the number of real solutions the equation has? What are the solutions to the quadratic equation 4(x +

    asked by chelsie on May 25, 2016
  3. Algebra

    Consider the equation 4x^2 – 16x + 25 = 0. (a) Show how to compute the discriminant, b^2 – 4ac, and then state whether there is one real-number solution, two different real-number solutions, or two different imaginary-number

    asked by Pamela on July 7, 2009
  4. Algebra

    Consider the equation: 4x^2 – 16x + 25 = 0 (a) Show how to compute the discriminant, b^2 – 4ac, and then state whether there is one real-number solution, two different real-number solutions, or two different imaginary-number

    asked by Tammie on July 8, 2009
  5. algebra

    Solve the equation using the zero-product property. -9n(5n-5) a. -1/9, 1 b. 0,1***** c. -1/9,-1 d. 0,-1 Use the quadratic formula to solve the equation. if necessary, round to the nearest hundredth. x^2-6=x a. x=2,3 b. x=-2,3

    asked by frank on May 14, 2018
  6. algebra

    It is possible for a quadratic equation to have no real-number solutions. Solve. t^2+10t+26=0

    asked by gina on July 6, 2011
  7. Algebra

    Compute the value of the discriminant and give the number of real solutions to the quadratic equation. 2x^2+5x-7=0 Discrimnant= number of real solutions=

    asked by Jessica on January 27, 2011
  8. Math

    Solve the given quadratic equation exactly using the quadratic formula. Write the solutions in its simplest form. Using a calculator, determine all irrational solutions to the nearest thousandth. Separate the solutions with a

    asked by Sam on September 23, 2014
  9. Math

    Solve the given quadratic equation exactly using the quadratic formula. Write the solutions in its simplest form. Using a calculator, determine all irrational solutions to the nearest thousandth. Separate the solutions with a

    asked by Jade on September 23, 2014
  10. Algebra

    I was given this answers to a problems I solved but I do not understand what she means hope you can help. Before attempting to solve this quadratic equation, determine how many solutions there will be for this quadratic equation.

    asked by Charly on June 2, 2011

More Similar Questions