if alpha and beta are the zeros of a polynomial x square - 7 x + K such that Alpha minus beta is equal to 1 then find the value of k
Answers
Answer : k = 12
Solution :
Given polynomial is
f(x) = x² - 7x + k
Since α and β are the zeroes of f(x), by the relation between zeroes and coefficients, we get
α + β = - (- 7/1) = 7
αβ = k
Given, α - β = 1
Squaring both sides, we get
(α - β)² = 1²
⇒ (α + β)² - 4αβ = 1
⇒ (- 7)² - 4k = 1
⇒ 49 - 4k = 1
⇒ 4k = 49 - 1 = 48
⇒ k = 48/4
∴ k = 12
K is 12.
K is 12.From the given quadratic equation, we can see that alpha + beta = 7. alpha - beta = 1 is given.
K is 12.From the given quadratic equation, we can see that alpha + beta = 7. alpha - beta = 1 is given.Let me use alpha as a and beta as b.
K is 12.From the given quadratic equation, we can see that alpha + beta = 7. alpha - beta = 1 is given.Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.
K is 12.From the given quadratic equation, we can see that alpha + beta = 7. alpha - beta = 1 is given.Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b = 1, we get
K is 12.From the given quadratic equation, we can see that alpha + beta = 7. alpha - beta = 1 is given.Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b = 1, we get4ab = 48
K is 12.From the given quadratic equation, we can see that alpha + beta = 7. alpha - beta = 1 is given.Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b
Let me use alpha as a and beta as b.
Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.
Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b = 1, we get
Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b = 1, we get4ab = 48
Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b = 1, we get4ab = 48ab = 12.
Let me use alpha as a and beta as b.(a+b)^2 -4ab = (a-b)^2.Substituting a+b = 7, a-b = 1, we get4ab = 48ab = 12.Therefore we can clearly see that k = 12