Question

Recall that a quadratic equation is written in the form of ax2 + bx + c = 0. For each equation below, identify a, b, and c.
b. Determine the value of the discriminant: b2 – 4ac.

Answer 1

@Michele_Laino

Answer 2

x would be b

Answer 3

hint:
I rewrite the first equation like below:
\((+1)\cdot x^2 + (-4) \cdot x+ (-5)=0\)
now, please compare such equation with the general equation: \(a \cdot x^2 + b \cdot x+ c=0\)

Answer 4

what is a b and c though

Answer 5

yes! what are:
\(a=...?\)
\(b=...?\)
\(c=...?\)

Answer 6

a=1
b=4
c=5?

Answer 7

please look at the coefficients inside the parentheses

Answer 8

sorry so b=-4
c=-5?

Answer 9

that's right!

Answer 10

now, the corresponding discriminant is:
\(\Delta= b^2-4ac= (-5)^2-[4 \cdot (+1) \cdot (-5)]=...?\)

Answer 11

oops.. sorry, here is the right formula:
\(\Delta= (-4)^2-[4 \cdot (+1) \cdot (-5)]=...?\)

Answer 12

so -4^2- 4 x1 x -5= 36 right?

Answer 13

that's right!
\(\Delta=16+20=36\)

Answer 14

cause 16- (-20)=36

Answer 15

yes!

Answer 16

what does the triangle stand for

Answer 17

\(\Delta\) is the greek letter \(delta\)

Answer 18

do I have to find square roots or no

Answer 19

cause the last part states determine whether the solution will be two real solutions, one real solution, or no real solution; two imaginary solutions.

Answer 20

I don't think, since the exercise asks for the coefficients and discriminant only.
In order to answer to last part, we have to establish the sign of \(\Delta\).
Now, since \(36>0\), then from the general theory we can state that there are two real solutions, and such solutions are different each from other

Answer 21

ok and how would you find the solutions

Answer 22

in order to find the solutions, I apply this formula:
\[\Large x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - \left( { - 4} \right) \pm \sqrt {36} }}{{2 \cdot 1}} = ...?\]

Answer 23

-1 and 5

Answer 24

yes! That's right!

Answer 25

ok Thank You!
I have A couple more of these Problems that I am going to do and Then Could you check them for me

Answer 26

ok!

Answer 27

8x^2+40x+50=0
a=8
b=40
c=50
40^2-4x8x50
1,600- 1,600=0
One solution but could you show me how i find that solution

Answer 28

@Michele_Laino

Answer 29

2x^2 +x+28=0
a=2
b=1
c=28
1-4x2x28= -911
no real solutions

Answer 30

@Michele_Laino

Answer 31

that's right!

Answer 32

more precisely, we have:
\(\Delta=1-56 \cdot 4=1-224=-223<0\)

Answer 33

how would I find the solution in the one equation?

Answer 34

@Michele_Laino

Answer 35

since \(\Delta<0\), in order to compute the solutions, we have to introduce the complex numbers.
Complex numbers are numbers \(z\) like below:
\(z=a+ib\) where \(a,\;b\) are real numbers, and \(i\) is such that \(i^2=-1\)

Answer 36

I got it Thank You! Could You help with a complex number thing?

Answer 37

ok!

Answer 38

Do you want me to post a new Question?

Answer 39

ok! :)