Question

X2+bx-32=0 x1=8 x2=? b=?​

Answer 1

х² + вх - 32 = 0

х1 * х2 = -32

8 * х2 = -32

х2= - 4

х1 + х2= - в

8 + (-4) = - в

в= - 4

х² - 4х - 32 = 0

ответ: х2= - 4;  в = - 4.

Answer 2

To find the values of x2 and b in the quadratic equation x^2 + bx - 32 = 0, given that x1 = 8, we can use the quadratic formula.

The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

Comparing this with our equation x^2 + bx - 32 = 0, we can see that a = 1, b = b, and c = -32.

Given that one of the solutions is x1 = 8, we can substitute these values into the quadratic formula to solve for b:

8 = (-b ± sqrt(b^2 - 4(1)(-32))) / (2(1))

Simplifying the equation:

8 = (-b ± sqrt(b^2 + 128)) / 2

Multiplying both sides by 2 to eliminate the denominator:

16 = -b ± sqrt(b^2 + 128)

Squaring both sides to eliminate the square root:

256 = b^2 + 128

Rearranging the equation:

b^2 = 256 - 128 b^2 = 128 b = ±√128

Therefore, the possible values for b are b = √128 and b = -√128.

To find the value of x2, we can substitute the value of b into the quadratic formula:

x2 = (-b ± sqrt(b^2 - 4ac)) / (2a)

Substituting the values a = 1, b = √128, and c = -32:

x2 = (-√128 ± sqrt((√128)^2 - 4(1)(-32))) / (2(1))

Simplifying the equation:

x2 = (-√128 ± sqrt(128 - 128)) / 2

x2 = (-√128 ± sqrt(0)) / 2

x2 = (-√128 ± 0) / 2

x2 = -√128 / 2 or x2 = √128 / 2

Simplifying further:

x2 = -8√2 or x2 = 8√2

Therefore, the values of x2 are x2 = -8√2 and x2 = 8√2, and the possible values of b are b = √128 and b = -√128.