If alpha and beta are the zeros f(x) = x square +5x +K
Such that alpha minus beta =1 find k
f(x) =xsquare -5x +K
Here,a=1 b=-5 c=k
Alpha+beta =-b/a
=5
Apha*beta =c/a
= K
Given alpha minus beta =1
Squaring both sides
We get,. ( Alpha minus beta)^2 =1^2
Alpha^2- 2alpha*beta+beta^2=1
(Alpha^2-2alpha*beta +beta^2)-4alpha*beta = 1
(Alpha + beta)^2-4alpha*beta=1
(5)^2- 4(k). = 1
25-4k=1
-4k=1-25
K=-24/-4
K=6
Thus the value of K is 6
To find the value of K, we need to use the conditions given: alpha and beta are the zeros of the quadratic equation f(x) = x^2 + 5x + K, and alpha - beta = 1.
The sum of the roots (zeros) of a quadratic equation is given by the formula:
Sum of roots = -b/a,
where a is the coefficient of the x^2 term (which is 1 in this case) and b is the coefficient of the x term (which is 5 in this case).
So, alpha + beta = -b/a = -5/1 = -5.
We also have the condition that alpha - beta = 1.
Given that alpha + beta = -5 and alpha - beta = 1, we can solve these equations simultaneously using the method of substitution or elimination.
Let's use the method of substitution:
We have the equations:
alpha + beta = -5 (Equation 1)
alpha - beta = 1 (Equation 2)
Let's solve Equation 2 for alpha:
alpha = beta + 1
Now substitute this value of alpha into Equation 1:
(beta + 1) + beta = -5
Simplifying the equation:
2beta + 1 = -5
2beta = -6
beta = -3
Now that we have the value of beta, we can substitute it back into Equation 2 to find alpha:
alpha - (-3) = 1
alpha + 3 = 1
alpha = -2
So, the values of alpha and beta are alpha = -2 and beta = -3.
Now let's find the value of K by substituting the values of alpha and beta into the quadratic equation f(x) = x^2 + 5x + K:
f(x) = (x - alpha)(x - beta)
f(x) = (x + 2)(x + 3)
Expanding:
f(x) = x^2 + 3x + 2x + 6
f(x) = x^2 + 5x + 6
Comparing this with the given equation f(x) = x^2 + 5x + K, we can see that K = 6.
Therefore, the value of K is 6.
For easier typing and reading I will let alpha be "a" and beta be "b"
so you want x^2 + 5x + K = 0
-- sum of roots = a+b = -5/1 = -5
-- product of roots = k(1) = ab
given:
a - b = 1
a + b = -5
add them:
2a = -4
a = -2, then b = -3
and k = ab = (-2)(-3) = 6
you had that, good job
since x = -b/2a ±√(b^2-4ac)/2a
a-b = 2±√(b^2-4ac)/2a = ±√(b^2-4ac)/a = 1
b^2-4ac = a^2
25-4k = 1
k = 6