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There are 5 quadratics below. Four of them have two distinct roots each. The other has only one distinct root; find the value of that root.
\( 4x^2 +16x - 9\\ 2x^2 + 80x + 400\\ x^2 - 6x - 9\\ 4x^2 - 12x + 9\\ {-x^2 + 14x + 49} \)

 May 31, 2019
 #1
avatar+8829 
+4

discriminant of first one   =   162 - 4(4)(-9)   =   400   ≠   0

 

discriminant of second one   =   802 - 4(2)(400)   =   3200   ≠   0

 

discriminant of third one   =   (-6)2 - 4(1)(-9)   =   72   ≠   0

 

discriminant of fourth one   =   (-12)2 - 4(4)(9)   =   0

 

discrimimant of fifth one   =   (14)2 - 4(-1)(49)   =   392   ≠   0

 

The fourth one is the quadratic with one distinct root.

 

4x2 - 12x + 9  =  0     when

 

x  =  \(\frac{-(-12)\pm\sqrt{(-12)^2-4(4)(9)}}{2(4)}\ =\ \frac{12\pm0}{8}\ =\ \frac{12}{8}\ =\ \frac32\)

 

Here's a graph of all of them: https://www.desmos.com/calculator/vrjvowhts0

 May 31, 2019
 #2
avatar+1015 
-3

Your on a rollllllll!!!

Nickolas  May 31, 2019

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