For what real number k does 3x^2 � k factor as
3(x � 2)(x + 2)?
Can someone please help with this question? I can not understand how it says to solve this in the book
Y = 3X^2 + K
Vertex Form:
Y = a(X -h)^2 + k
= 3(X -0)^2 + K
Y = 3(X-2)(X+2),
= 3(X^2 -4)
= 3X^2 - 12
h = -b / 2a = 0 / 6 = 0.
K = Yv = 3*0^2 -12 = -12. V(0 , -12).
Let's use a positive 2 in the 1st
parenthesis of our factored Eq.
Y = 3(X + 2 )(X + 2) = 3(X + 2)^2 =
3(X^2 + 4X + 4) = 3X^2 + 12X + 12.
h = Xv = -b / 2a = -12 / 6 = -2.
This value of h does not agree with
our given Eq in which h = 0.
Therefore, -2 is the correct choice.
To solve this question, we need to compare the given expression 3x^2 � k with the factored form 3(x � 2)(x + 2).
In the factored form, we have two factors: (x - 2) and (x + 2). To get the expression in the factored form, we need to expand the factors: (x - 2)(x + 2) = x^2 - 4.
Now, let's compare the expanded form with the given expression:
x^2 - 4 � k
This tells us that the expression x^2 - 4 is equivalent to 3x^2 � k.
To make the comparison easier, let's rewrite the expression x^2 - 4 in terms of the variable k: (3/1)(x^2 - 4)
Now we have:
(3/1)(x^2 - 4) � k
Comparing the coefficients of the x^2 term on both sides, we have:
3/1 = 3
Thus, we can conclude that the real number k is equal to 3.