Answer :
To determine the values of [tex]\( n \)[/tex] for which the quadratic equation [tex]\( M^2 + nM + n = 0 \)[/tex] has zero real roots, we should analyze the discriminant of the quadratic equation.
A quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] has a discriminant [tex]\( \Delta \)[/tex] given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the equation [tex]\( M^2 + nM + n = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = n \)[/tex]
- [tex]\( c = n \)[/tex]
So, the discriminant [tex]\( \Delta \)[/tex] is:
[tex]\[ \Delta = n^2 - 4 \cdot 1 \cdot n \][/tex]
[tex]\[ \Delta = n^2 - 4n \][/tex]
For the quadratic equation to have zero real roots, the discriminant must be less than zero ([tex]\( \Delta < 0 \)[/tex]):
[tex]\[ n^2 - 4n < 0 \][/tex]
To solve this inequality, factor the quadratic expression:
[tex]\[ n(n - 4) < 0 \][/tex]
This inequality represents the values of [tex]\( n \)[/tex] where the product of [tex]\( n \)[/tex] and [tex]\( (n - 4) \)[/tex] is negative. The solutions to the inequality can be found by considering the intervals where the product switches signs. The critical points are [tex]\( n = 0 \)[/tex] and [tex]\( n = 4 \)[/tex].
Plotting these points on a number line and testing intervals:
- For [tex]\( n < 0 \)[/tex]: both [tex]\( n \)[/tex] and [tex]\( (n - 4) \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < n < 4 \)[/tex]: [tex]\( n \)[/tex] is positive and [tex]\( (n - 4) \)[/tex] is negative, so their product is negative.
- For [tex]\( n > 4 \)[/tex]: both [tex]\( n \)[/tex] and [tex]\( (n - 4) \)[/tex] are positive, so their product is positive.
Therefore, the inequality [tex]\( n(n - 4) < 0 \)[/tex] holds in the interval:
[tex]\[ 0 < n < 4 \][/tex]
Thus, for the quadratic equation [tex]\( M^2 + nM + n = 0 \)[/tex] to have zero real roots, the values of [tex]\( n \)[/tex] must lie within the interval:
[tex]\[ (0, 4) \][/tex]
So, the values of [tex]\( n \)[/tex] for which the quadratic [tex]\( M^2 + nM + n = 0 \)[/tex] has zero real roots are in the interval [tex]\( (0, 4) \)[/tex].
A quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] has a discriminant [tex]\( \Delta \)[/tex] given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the equation [tex]\( M^2 + nM + n = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = n \)[/tex]
- [tex]\( c = n \)[/tex]
So, the discriminant [tex]\( \Delta \)[/tex] is:
[tex]\[ \Delta = n^2 - 4 \cdot 1 \cdot n \][/tex]
[tex]\[ \Delta = n^2 - 4n \][/tex]
For the quadratic equation to have zero real roots, the discriminant must be less than zero ([tex]\( \Delta < 0 \)[/tex]):
[tex]\[ n^2 - 4n < 0 \][/tex]
To solve this inequality, factor the quadratic expression:
[tex]\[ n(n - 4) < 0 \][/tex]
This inequality represents the values of [tex]\( n \)[/tex] where the product of [tex]\( n \)[/tex] and [tex]\( (n - 4) \)[/tex] is negative. The solutions to the inequality can be found by considering the intervals where the product switches signs. The critical points are [tex]\( n = 0 \)[/tex] and [tex]\( n = 4 \)[/tex].
Plotting these points on a number line and testing intervals:
- For [tex]\( n < 0 \)[/tex]: both [tex]\( n \)[/tex] and [tex]\( (n - 4) \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < n < 4 \)[/tex]: [tex]\( n \)[/tex] is positive and [tex]\( (n - 4) \)[/tex] is negative, so their product is negative.
- For [tex]\( n > 4 \)[/tex]: both [tex]\( n \)[/tex] and [tex]\( (n - 4) \)[/tex] are positive, so their product is positive.
Therefore, the inequality [tex]\( n(n - 4) < 0 \)[/tex] holds in the interval:
[tex]\[ 0 < n < 4 \][/tex]
Thus, for the quadratic equation [tex]\( M^2 + nM + n = 0 \)[/tex] to have zero real roots, the values of [tex]\( n \)[/tex] must lie within the interval:
[tex]\[ (0, 4) \][/tex]
So, the values of [tex]\( n \)[/tex] for which the quadratic [tex]\( M^2 + nM + n = 0 \)[/tex] has zero real roots are in the interval [tex]\( (0, 4) \)[/tex].