To determine for which values of [tex]\( n \)[/tex] the equation [tex]\( 3u^2 - 6u = n + 1 \)[/tex] has two real solutions, follow these detailed steps:
1. Rewrite the equation in standard quadratic form:
[tex]\[
3u^2 - 6u - (n + 1) = 0
\][/tex]
This quadratic equation is in the form [tex]\( au^2 + bu + c = 0 \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -(n + 1) \)[/tex].
2. Recall the condition for a quadratic equation to have two distinct real solutions:
A quadratic equation [tex]\( au^2 + bu + c = 0 \)[/tex] has two distinct real solutions if its discriminant is greater than zero. The discriminant [tex]\( \Delta \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
3. Calculate the discriminant for the given quadratic equation:
[tex]\[
\Delta = (-6)^2 - 4 \cdot 3 \cdot -(n + 1)
\][/tex]
Simplify the expression:
[tex]\[
\Delta = 36 + 12(n + 1)
\][/tex]
[tex]\[
\Delta = 36 + 12n + 12
\][/tex]
[tex]\[
\Delta = 48 + 12n
\][/tex]
4. Set up the inequality for the discriminant to be greater than zero:
[tex]\[
48 + 12n > 0
\][/tex]
5. Solve the inequality for [tex]\( n \)[/tex]:
[tex]\[
12n > -48
\][/tex]
[tex]\[
n > -4
\][/tex]
Therefore, the equation [tex]\( 3u^2 - 6u = n + 1 \)[/tex] will have two real solutions for values of [tex]\( n \)[/tex] that satisfy:
[tex]\[
n > -4
\][/tex]
