How do you find the discriminant and how many and what type of solutions does x^2+3x+7=0x2+3x+7=0 have?

1 Answer
Zor Shekhtman
May 10, 2015

Let's get some theoretical background first.
Assume you have to solve a general quadratic equation
ax^2+bx+c=0ax2+bx+c=0
You have no recollection of the formula for its solution (like most people). How should you solve it?
Let's apply the logic.

If your equation looks like this
alpha (x+ beta)^2 + gamma = 0α(x+β)2+γ=0,
the solution is easy:

  1. You subtract gammaγ from both sides, getting
    alpha(x+beta)^2=-gammaα(x+β)2=−γ
  2. Divide by alphaα both sides, getting
    #(x+beta)^2=-gamma/alpha
  3. Extract a square root from both side using positive and negative values of a square root, getting
    x+beta=sqrt(-gamma/alpha)x+β=√−γα
  4. Subtract betaβ from both sides, getting
    x_(1,2)=+-sqrt(-gamma/alpha)-betax1,2=±√−γα−β

Our general equation is not of the form we can easily solve, as above, but can be transformed into this form.
Let's equate two forms and find the necessary alpha, beta, gammaα,β,γ coefficients:
ax^2+bx+c=alpha(x+beta)^2+gammaax2+bx+c=α(x+β)2+γ

Open the parenthesis on the right:
ax^2+bx+c=alpha x^2 + 2 alpha beta x + alpha beta^2 + gammaax2+bx+c=αx2+2αβx+αβ2+γ

Now you can determine alpha, beta, gammaα,β,γ coefficients by comparing the coefficient at xx in different degrees:
a = alphaa=α (coefficient at x^2x2)
b = 2 alpha betab=2αβ (coefficient at xx)
c = alpha beta ^2 + gammac=αβ2+γ (free coefficient)
from which follows:
alpha = aα=a
beta = b/(2 alpha) = b/(2a)β=b2α=b2a
gamma = c - alpha beta ^2 = c - a b^2/(4a^2) = c - b^2/(4a) = (4ac-b^2)/(4a)γ=c−αβ2=c−ab24a2=c−b24a=4ac−b24a

Now we use the formula that expressed the solution in terms of alpha, beta, gammaα,β,γ, substituting these coefficients with their representation in terms of a, b, ca,b,c.

x_(1,2)=+-sqrt(-gamma/alpha)-beta = +=sqrt((-4ac+b^2)/(4a))-b/(2a) = [-b+-sqrt(b^2-4ac)]/(2a)x1,2=±√−γα−β=+=√−4ac+b24a−b2a=−b±√b2−4ac2a

The most involved part of this expression is under the square root, it's called discriminant of this quadratic equation:
D = b^2-4acD=b2−4ac

If D=b^2-4ac>0D=b2−4ac>0, there are two distinct real solution to this quadratic equation because the square root of the positive number has two distinct real values.

If D=b^2-4ac=0D=b2−4ac=0, there is only one real solution since the square root equals to 0 and our +-± operation renders always 00.

Finally, if D=b^2-4ac < 0D=b2−4ac<0, there are no real solution because there are no real values for a square root of a negative number.

In a particular case of this problem with an equation
x^2+3x+7=0x2+3x+7=0
the coefficients are:
a=1, b=3, c=7a=1,b=3,c=7
The discriminant D=b^2-4ac=3^2-4*1*7=-19D=b2−4ac=32−4⋅1⋅7=−19
It's negative. Therefore, there are no real solutions.

A more detailed explanation of topics touched in this answer can be found on Unizor by following the menu items Algebra - Quadratic Equations.

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