SOLUTION: find discriminant of the following quadratic equation and examine the nature of real roots(if they exist) 7y^2 + 4y +5 = 0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: find discriminant of the following quadratic equation and examine the nature of real roots(if they exist) 7y^2 + 4y +5 = 0      Log On


   



Question 935880: find discriminant of the following quadratic equation and examine the nature of real roots(if they exist) 7y^2 + 4y +5 = 0
Found 2 solutions by Edwin McCravy, Theo:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


Therefore, since the discriminant is what is under the 
             _____
square root √      in the quadratic formula,

y+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29+,
                                               _____  
The negative number -124 being underneath the √ 
will cause there to be an imaginary root, and since
the quadratic equation can be written as

-b%2F%282a%29+%2B+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29+, and -b%2F%282a%29+-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29+,

There will be two conjugate imaginary/complex roots, neither of
which is real.  There are no real roots.

Edwin

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the discriminant is b^2 - 4ac

your equation of 7y^2 + 4y +5 = 0 is in standard form of ax^2 + bx + c = 0.

just change your y to an x and you can match the standard form.

your equation becomes:

7x^2 + 4x + 5 = 0

a = 7
b = 4
c = 5

b^2 - 4ac = 4^2 - 4 * 7 * 5 which becomes:

b^2 - 4ac = 16 - 140 which becomes:

b^2 - 4ax = -124

your discriminant is negative, so your solution will not be real, i.e. it will include an imaginary component.

the rules are:

if the discriminant is positive, then the roots are real.

if the discriminant is negative, then the roots will not be real.

you can solve the equation using the quadratic formula to see what the solution will be.

the quadratic formula equation is x = (-b plus or minus sqrt(b^2 - 4ac)) / (2a)

solve this equation and you will find that your roots are:

(-4 +/- sqrt(-124)) / 14 which becomes:

(-4 +/- sqrt(124) * i) / 14

you get two roots.

first root is -4/14 + sqrt(124) * i

second root is -4/14 - sqrt(124) * i

these could probably be simplified further, but i didn't do that since it was not necessary to show you that the roots are not real.

if you were solving for the roots, you would probably have had to simplify it as much as possible.

the roots are not real because they have an imaginary component.
that imaginary component is sqrt(-124) which becomes sqrt(124) * sqrt(-1) which becomes sqrt(124) * i.

this is because i is equal to sqrt(-1).

since the roots are not real, the graph of the equation will not cross the x-axis at those roots.

the graph of the equation only crosses the x-axis when the roots are real.

the graph of the equation of y = 7x^2 + 4x + 5 is shown below:

+graph+%28600%2C600%2C-5%2C5%2C-10%2C10%2C7x%5E2+%2B+4x+%2B+5%29