Quadratic Equations

Question

How many real-number solutions does the equation have?

-7x^2+6x+3=0

 

How many real-number solutions does the equation have?

0=3x^2+18x+27

Answer 1

The way to determine the nature of the solutions is to find the "discriminant" of the quadratic polynomial, which is equal to b² - 4ac.  If the discriminant is positive, there are 2 real solutions.  If it is 0, there is 1 real solution.  If it is negative there are no real solutions and 2 complex solutions.

 

From that, we can derive a simpler test that covers some, but not all cases.

 

If 4ac is negative, then b² - 4ac MUST be positive, because b² must be positive and you'd be SUBTRACTING a negative from a positive, and the result must be positive.  And since 4 is positive, all you need to test is whether ac is negative.  If it IS, then there are 2 real roots.  If it is NOT, then you have to calculate the full discriminant.

 

-7x² + 6x + 3 = 0 ... ac = (-7)*3 = -21, which is negative.  So there are two real solutions.

 

0 = 3x² + 18x + 27 ... ac = 3*27 = 81, which is positive.  So you have to calculate the discriminant, b² - 4ac, which is 18² - 4*3*27 = 324 - 324 =  0.  So there is one real solution.

 

 

 

Answer 2

The number and nature of roots to quadratic equations can be determined with the discriminant of the quadratic formula: b² - 4ac

 

If b² - 4ac < 0, two imaginary roots

If b² - 4ac = 0, two equal real roots

If b² - 4ac > 0, two unequal real toots

 

For the first equation

 

6² - 4(-7)(3)

36 + 84

120

 

Two unequal real roots.

 

Do the same for the second equation.