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Sagot :
To determine the number of real-number solutions for the quadratic equation [tex]\( 0 = 3x^2 - 5x - 5 \)[/tex], we need to analyze the discriminant of the equation. A quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex] has a discriminant given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
For the equation [tex]\( 0 = 3x^2 - 5x - 5 \)[/tex]:
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = -5 \)[/tex]
- [tex]\( c = -5 \)[/tex]
The discriminant [tex]\( \Delta \)[/tex] can be calculated as:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Inserting the given values:
[tex]\[ \Delta = (-5)^2 - 4(3)(-5) \][/tex]
[tex]\[ \Delta = 25 + 60 \][/tex]
[tex]\[ \Delta = 85 \][/tex]
Next, we interpret the discriminant [tex]\( \Delta \)[/tex]:
- If [tex]\( \Delta > 0 \)[/tex], the equation has two distinct real solutions.
- If [tex]\( \Delta = 0 \)[/tex], the equation has exactly one real solution.
- If [tex]\( \Delta < 0 \)[/tex], the equation has no real solutions (the solutions are complex).
Since [tex]\( \Delta = 85 \)[/tex], which is greater than 0, the quadratic equation [tex]\( 0 = 3x^2 - 5x - 5 \)[/tex] has two distinct real solutions.
Therefore, the answer is:
[tex]\[ \boxed{\text{two solutions}} \][/tex]
For the equation [tex]\( 0 = 3x^2 - 5x - 5 \)[/tex]:
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = -5 \)[/tex]
- [tex]\( c = -5 \)[/tex]
The discriminant [tex]\( \Delta \)[/tex] can be calculated as:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Inserting the given values:
[tex]\[ \Delta = (-5)^2 - 4(3)(-5) \][/tex]
[tex]\[ \Delta = 25 + 60 \][/tex]
[tex]\[ \Delta = 85 \][/tex]
Next, we interpret the discriminant [tex]\( \Delta \)[/tex]:
- If [tex]\( \Delta > 0 \)[/tex], the equation has two distinct real solutions.
- If [tex]\( \Delta = 0 \)[/tex], the equation has exactly one real solution.
- If [tex]\( \Delta < 0 \)[/tex], the equation has no real solutions (the solutions are complex).
Since [tex]\( \Delta = 85 \)[/tex], which is greater than 0, the quadratic equation [tex]\( 0 = 3x^2 - 5x - 5 \)[/tex] has two distinct real solutions.
Therefore, the answer is:
[tex]\[ \boxed{\text{two solutions}} \][/tex]
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