Hello, pboland!
Identify the coefficients, and compute the discriminant, for each quadratic equation in the table.
Code:
Equation a b c Discriminant
x² - 1 = 0 1 1 0 1
x² + 1 = 0 1 1 0 1
1 - 9x² = 0 1 9 0 1
x² + 3x = 0 1 3 0 1
I used \(\displaystyle b^2\,-\,4ac\) for the discriminant formula.
Are these correct?
. . . sorry, no -- all wrong!
\(\displaystyle a,\,b,\,c\) are the coefficients of: \(\displaystyle ax^2\,+\,bx\,+\,c\)
So \(\displaystyle a\) is the coefficient of the \(\displaystyle x^2\)-term,
no matter where it is written.
\(\displaystyle \;\:\:b\) is the coefficient of the \(\displaystyle x\)-term.
\(\displaystyle \;\:\:c\) is the constant term.
The first one is actually: \(\displaystyle \,1\cdot x^2\,+\,0\cdot x\,-\,1\;=\;0\)
\(\displaystyle \;\;\)Hence: \(\displaystyle a\,=\,1,\;\;b\,=\,0,\;\;c\,=\,-1\)
\(\displaystyle \;\;\)and: \(\displaystyle \,D\:=\:0^2\,-\,4(1)(-1)\;=\;4\)
The second is: \(\displaystyle \,1\cdot x^2\,+\,0\cdot x\,+\,1\;=\;0\)
\(\displaystyle \;\;\)Hence: \(\displaystyle \,a\,=\,1,\;\;b\,=\,0,\;\;c\,=\,1\)
\(\displaystyle \;\;\)and: \(\displaystyle \,D\;=\;0^2\,-\,4(1)(1)\;=\;-4\)
The third is: \(\displaystyle \,-9\cdot x^2\,+\,0\cdot x\,+\,1\;=\;0\)
\(\displaystyle \;\;\)Hence: \(\displaystyle \,a\,=\,-9,\;\;b\,=\,0,\;\;c\,=\,1\)
\(\displaystyle \;\;\)and: \(\displaystyle \,D\;=\;0^2\,-\,4(-9)(1)\;=\;36\)
The fourth is: \(\displaystyle \,1\cdot x^2\,+\,3\cdot x\,+\,0\;=\;0\)
\(\displaystyle \;\;\)Hence: \(\displaystyle a\,=\,1,\;\;b\,=\,3,\;\;c\,=\,0\)
\(\displaystyle \;\;\)and: \(\displaystyle \,D\;=\;3^2\,-\,4(1)(0)\;=\;9\)
Edit: Too fast for me, Steve!
