<em>Step #1: </em>
Make sure the equation is in the form of [ Ax² + Bx + C = 0 ].

Yours is already in that form.
A = 1
B = 2
C = -2

<em>Step #2:</em>
The 'discriminant' for that equation is [ B² - 4 A C ].
That's all there is to it, but it can tell you a lot about the roots of the equation.

-- If the discriminant is zero, then the left side of the equation is a perfect square,
and both roots are equal.

-- If the discriminant is greater than zero, the the roots are real and not equal.

-- If the discriminant is less than zero, then the roots are complex numbers.

The discriminant of your equation is [ B² - 4 A C ] = 2² - 4(1)(-2) = 4 + 8 = 12

Your equation has two real, unequal roots.

4 0

3 years ago

Read 2 more answers

Can anyone do question 25 part c? :)

inna [77]

$\begin{cases}x^3\qquad\qquad\ \ \ when\ x\geq3\\5\qquad\qquad\quad \ when\ 0\ \textless \ x\ \textless \ 3\\x^2-x+2\quad\ when\ x\leq0 \end{cases}\\\\\\ (a)\\ f(0)\iff x=0\implies f(x)=x^2-x+2\\\\f(0)=0^2-0+2=2\\\\(b)\\ \{f(2)\iff x=2;\ \ f(1)\iff x=1\}\implies f(x)=5\\\\ f(2)=5\quad\wedge\quad f(1)=5\\\\ f(2)-f(1)=5-5=0$

$(c)\\ \big\forall \limits_{ n\in R}\ n^2\geq 0\ \ \Rightarrow\ \ \big\forall \limits_{ n\in R}\ (-n^2)\leq 0 \ \ \Rightarrow\ \ x\leq0\ \ \Rightarrow\ \ f(x)=x^2-x+2\\\\f(-n^2)=(-n^2)^2-(-n^2)+2=n^4+n^2+2$

5 0

4 years ago

When two events are mutually exclusive, why is P(A and B)=0?

kati45 [8]

Answer:

Step-by-step explanation:

"mutually exclusive" means "cannot happen at the same time." There's no "union" (no A ∩ B). The probability of this "union" is zero.

5 0

3 years ago