Question

Will someone please help me?!?

Justify if completing the square is a good method for solving when the Discriminant is negative. Use any of your three functions as an example and respond in complete sentences.

Answer 1

The function I created I want to use is -->
\[f(x)=(x-3)(x-2)\]

Answer 2

@mathmale
@austinL
@maxdrizner

Answer 3

@phi can you please help I don't understand

Answer 4

i honestly dont know how to do this,,, sorry

Answer 5

Thank you anyway.

Answer 6

The question says
***
Justify if completing the square is a good method for solving when the Discriminant is negative.
***

the discriminant is b^2 - 4ac
where you a,b and c are the coefficients of the quadratic, when it is written in standard form:
\[ y = a x^2 + bx + c \]

Answer 7

I already know the standard form of my first function is
\[y=x ^{2}-5x+6\]

Answer 8

Kid STOP! I am trying to learn something.

Answer 9

I blocked him, sorry

Answer 10

what is the discriminant of your function ?

Answer 11

I'm not sure what the discriminant is.

Answer 12

the discriminant is b^2 - 4*a*c
in your case a=1 (number in front of x^2)
b = -5 and c= 6

Answer 13

okay I'm following you.

Answer 14

you would get 1 for the discriminant.

I don't see how that helps us, because they are asking about a negative discriminant.
for example, find the discriminant of
\[ f(x) = x^2 -2x + 2\]

Answer 15

can you find the discriminate of
\[ f(x) = x^2 -2x + 2 \]
a=1, b= -2 and c=2

Answer 16

if you using completing the square on
\[ x^2 -2x + 2 = 0 \\ x^2 -2x = -2 \\ x^2 -2x +(-2/2)^2 = -2 +(-2/2)^2 \\
x^2 -2x + 1 = -1 \\
(x-1)^2 = -1
\]
now take the square root of both sides
\[ \sqrt{(x-1)^2} = \sqrt{-1} \\
x-1 = \sqrt{-1} \\
x= 1 ± \sqrt{-1}
\]
That is the correct answer, but
if you don't know about complex numbers, then you don't know how to take the square root of a negative number.

Answer 17

I know \[\sqrt{-1} = i\]

Answer 18

@phi

Answer 19

So that was an example of using "complete the square" to solve a quadratic whose discriminant is negative.

If none of your examples has a negative discriminant maybe you should add that example to your list.

Answer 20

okay thank you. I really appreciate you helping me.

Answer 21

can you help me with one more thing??
I need help finding the solutions of that same function.
\[g(x)=(x-\sqrt{3})(x+\sqrt{3})\]

Answer 22

\[x=\sqrt{3},-\sqrt{3}\]

Answer 23

\[g(\sqrt{3})=(x-\sqrt{3})(x+\sqrt{3})\]

Answer 24

\[g=\frac{ (x-\sqrt{3})(x+\sqrt{3}) }{ \sqrt{3} }\]

Answer 25

When you find the solutions of a function, you need an equation.
In other words if you have a function g(x)
you could ask, "what x makes g(x) = 0 ? " or "what x makes g(x) = 3 ?" , etc.

When you solve for the "roots of an equation" that means solve
g(x) = 0

Answer 26

***
I need help finding the solutions of that same function.
\[ g(x)=(x-\sqrt{3})(x+\sqrt{3}) \]
***

that is not an equation, that is a Definition... it explains that g(x) is short-hand for the stuff on the right side of the = sign.

if you had, for example
\[ (x-\sqrt{3})(x+\sqrt{3}) = 0 \]
then the solutions would be
x= -sqrt(3) and x= + sqrt(3)

Answer 27

notice that finding the solutions is very easy, because f(x) is in factored form. If you started with
\[ g(x) = x^2 -3 \]
and had to solve for
\[ x^2 -3 = 0 \]
you normally would have to factor the left-hand side.
But in this case, it is still easy to solve:
\[ x^2 = 3 \\
x = ± \sqrt{3} \]

Answer 28

btw,
\( g(\sqrt{3} )\) is short-hand for replace "x" with \( sqrt{3} \) in the definition of g(x)
you would say:
\[ g(x)=(x-\sqrt{3})(x+\sqrt{3}) \\
g(\sqrt{3} )=(\sqrt{3} -\sqrt{3})(\sqrt{3} +\sqrt{3}) \\
g(\sqrt{3} )= 0 \cdot 2 \sqrt{3}\\
g(\sqrt{3} )= 0
\]
you would also get 0 if you used - sqrt(3)