X1*X2=-16 X1+X2=11,25 X1=? X2=?
Ответ нейросети
19 марта 2025 17:19
To solve the system of equations:
- ( X_1 \times X_2 = -16 ) (Equation 1)
- ( X_1 + X_2 = 11.25 ) (Equation 2)
We can express ( X_2 ) in terms of ( X_1 ) using Equation 2:
[ X_2 = 11.25 - X_1 ]
Now, we can substitute this expression for ( X_2 ) into Equation 1:
[ X_1 \times (11.25 - X_1) = -16 ]
Expanding this gives:
[ 11.25X_1 - X_1^2 = -16 ]
Rearranging the equation leads to:
[ X_1^2 - 11.25X_1 - 16 = 0 ]
Now we can solve this quadratic equation using the quadratic formula:
[ X_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
In this case, ( a = 1 ), ( b = -11.25 ), and ( c = -16 ):
[ X_1 = \frac{-(-11.25) \pm \sqrt{(-11.25)^2 - 4(1)(-16)}}{2(1)} ]
Calculating the discriminant:
[ (-11.25)^2 = 126.5625 ]
[ 4 \times 1 \times -16 = -64 ]
So:
[ b^2 - 4ac = 126.5625 + 64 = 190.5625 ]
Now substituting back in:
[ X_1 = \frac{11.25 \pm \sqrt{190.5625}}{2} ]
Calculating the square root:
[ \sqrt{190.5625} \approx 13.8 ]
Now substituting back:
[ X_1 = \frac{11.25 \pm 13.8}{2} ]
We have two options for ( X_1 ):
- ( X_1 = \frac{11.25 + 13.8}{2} \approx \frac{25.05}{2} \approx 12.525 )
- ( X_1 = \frac{11.25 - 13.8}{2} \approx \frac{-2.55}{2} \approx -1.275 )
Now, we calculate ( X_2 ) for each ( X_1 ):
For ( X_1 \approx 12.525 ): [ X_2 = 11.25 - 12.525 \approx -1.275 ]
For ( X_1 \approx -1.275 ): [ X_2 = 11.25 - (-1.275) \approx 12.525 ]
Thus, the pairs ((X_1, X_2)) are:
- ( X_1 \approx 12.525 ) and ( X_2 \approx -1.275 )
- ( X_1 \approx -1.275 ) and ( X_2 \approx 12.525 )
Final answers:
- ( X_1 \approx 12.525, X_2 \approx -1.275 )
- or
- ( X_1 \approx -1.275, X_2 \approx 12.525 )
Понравился ответ?
Задай свой вопрос
Напиши любую задачу или вопрос, а нейросеть её решит



