Question

X^2+kx+7=0, x1-x2=1, k-?

Answer 1

X²+kx+7=0

x1+x2=-k
x1×x2=7


{x1+x2=-k
{x1-x2=1

x1+x2-x1-(-x2)=-k-1
x1+x2-x1+x2=-k-1
2x2=-k-1|÷2

x2=-(k+1)/2

x1+x2+x1-x2=-k+1
2x1=1-k|÷2

x1=(1-k)/2

x1×x2=7

((1-k)/2)×(-(k+1)/2)=7|×(-2)
(1-k)×(1+k)=-14
1-k²=-14
-k²=-14-1
-k²=-15|×(-1)
k²=15

k1=-√15
k2=√15

Answer 2

We can use the fact that the difference between the roots of a quadratic equation of the form ax^2 + bx + c = 0 is given by:

x1 - x2 = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, we have the equation:

x^2 + kx + 7 = 0

Comparing this to the standard form of a quadratic equation ax^2 + bx + c = 0, we see that a = 1, b = k, and c = 7. Thus, we have:

x1 - x2 = (-k ± sqrt(k^2 - 4(1)(7))) / (2(1)) = (-k ± sqrt(k^2 - 28)) / 2

We also know that x1 - x2 = 1. Therefore, we can set up the following equation:

1 = (-k ± sqrt(k^2 - 28)) / 2

Multiplying both sides by 2 and rearranging, we get:

2 = -k ± sqrt(k^2 - 28)

Squaring both sides, we get:

4 = k^2 - 28

Adding 28 to both sides, we get:

k^2 = 32

Taking the square root of both sides (note that we are only interested in the positive square root, since k is a coefficient in a quadratic equation), we get:

k = sqrt(32)

Simplifying this expression, we get:

k = 4sqrt(2)

Therefore, the value of k is 4sqrt(2).