How many real number solutions? #-7x^2+6x+3=0#

2 Answers
Bill Jorgensen
May 28, 2018

discriminant = 120

two real irrational solutions.

Explanation:

#-7x^2+6x+3#

To solve this use the discriminant:

#ax^2+bx+c#

discriminant #=b^2-4ac#

#=6^2-4*-7*3#

#=36-(-84)#

#=120#

discriminant > 0: two real solutions.

discriminant = 0: one real solutions, bounce or double solution.

discriminant < 0: two imaginary solutions.

discriminant = perfect square: solution is rational

Douglas K.
May 28, 2018

#d = 120#, therefore, there are two distinct real roots.

Explanation:

One can find the number of real solutions of a quadratic of the form, #y = ax^2+bx+c#, by computing the determinant, #d = b^2-4ac#

If #d < 0#, then there are no real roots.
If #d=0#, then there is one real root (called a repeated root).
If #d >0#, then there are two distinct real roots:

Given: #-7x^2+6x+3=0#

#d = 6^2-4(-7)(3)#

#d = 120#, therefore, there are two distinct real roots.

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