Question

A) x²+1=0; 6)x²-8x+65=0. : ITV

Answer 1

Ответ:

Объяснение:

x²+1=0; x²=-1; x=√(-1)

Из отрицательного числа корень не извлекается, уравнение не имеет решений.

x²-8x+65=0; D=64-260=-196

При D<0 уравнение не имеет решений.

Answer 2

A) x²+1=0 has no real solutions since the square of any real number is non-negative, and therefore x²+1 can never equal zero.

For x²+1=0, we can solve it by considering complex numbers. Let's define the imaginary unit i as √(-1), so that i²=-1. Then we have:

x²+1 = 0 x² = -1 x = ±√(-1) x = ±i

Therefore, the solutions to x²+1=0 are x = i and x = -i.

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6. To find the interval of real values of x for which x²-8x+65=0 is true, we can use the quadratic formula:

x = (-b ± √(b²-4ac)) / 2a

In this case, a=1, b=-8, and c=65. Substituting these values into the formula, we get:

x = (8 ± √((-8)²-4(1)(65))) / 2(1) x = (8 ± √(-156)) / 2

Since the discriminant (b²-4ac) is negative, the square root of the discriminant is an imaginary number. Therefore, the solutions to the equation x²-8x+65=0 are also complex numbers, and there are no real solutions.

Note that we could also see this by observing that the quadratic function x²-8x+65 is a parabola that opens upwards (since the coefficient of x² is positive), and its vertex lies above the x-axis (since the x-coordinate of the vertex is -b/2a=4 and the y-coordinate is f(4)=1). Therefore, the function does not intersect the x-axis and has no real roots.