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Sagot :
There is a maximum possible of 2 solutions since this is a polynomial (with a power of 2).
I'm not sure what the easiest way is to determine whether the solutions are real or complex, but I do know of a way that is simple enough to be performed by you . . .
Use the quadratic formula (this calculates the roots of a quadratic equation):
x₁ = [-b + √(b² - 4ac)]/(2a)
x₂ = [-b - √(b² - 4ac)]/(2a)
where:
y = ax² + bx + c
y = 3x² - 5x - 5
so . . .
a = +3
b = -5
c = -5
plug & chug and you get . . .
x₁ = [-3 + √(85)]/(6)
x₂ = [-3 - √(85)]/(6)
since the value under the square root sign is positive (i.e. √(85)), the roots x₁ & x₂ will be positive, and thus they'll have real solutions (note: if it was √(-85), the solution would involve a complex number, so it would not be a real solution)
. . . answer is . . . (b) two
I'm not sure what the easiest way is to determine whether the solutions are real or complex, but I do know of a way that is simple enough to be performed by you . . .
Use the quadratic formula (this calculates the roots of a quadratic equation):
x₁ = [-b + √(b² - 4ac)]/(2a)
x₂ = [-b - √(b² - 4ac)]/(2a)
where:
y = ax² + bx + c
y = 3x² - 5x - 5
so . . .
a = +3
b = -5
c = -5
plug & chug and you get . . .
x₁ = [-3 + √(85)]/(6)
x₂ = [-3 - √(85)]/(6)
since the value under the square root sign is positive (i.e. √(85)), the roots x₁ & x₂ will be positive, and thus they'll have real solutions (note: if it was √(-85), the solution would involve a complex number, so it would not be a real solution)
. . . answer is . . . (b) two
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