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If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = ?

By BYJU'S Exam Prep

Updated on: October 17th, 2023

(1/α + 1/β) is -3.

Let us see the steps which we will be using to solve the given problem.

Step 1: Use the property that the sum of the roots of a quadratic equation ax^2 + bx + c = 0 is equal to -b/a.
Step 2: Identify the coefficients of the quadratic equation. In this case, a = 1 and b = 6.
Step 3: Apply the property to find the sum of the roots: Sum of the roots = -b/a = -6/1 = -6.
Step 4: Determine the product of the roots (α × β) by dividing the constant term (c) by the coefficient of the x^2 term (a).
Step 6: Calculate the product of the roots: Product of the roots = c/a = 2/1 = 2.
Step 7: Use (1/α + 1/β): (1/α + 1/β) = (α + β)/(α× β)

Table of content

1. If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = ?
2. If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = -3

If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = ?

Solution:

To find the value of (1/α + 1/β), the quadratic polynomial is x^2 + 6x + 2, and the roots are α and β.

So, the sum of the roots is -b/a, where a = 1 and b = 6:

Sum of the roots = -b/a = -6/1 = -6.

Now, we need to find the value of (1/α + 1/β). To do this, we can use the fact that (1/α + 1/β) = (α + β)/(α × β) by taking the reciprocal of each root and then finding their sum.

From the quadratic equation, we know that the product of the roots (α × β) is equal to c/a, where a = 1 and c = 2:

Product of the roots = c/a = 2/1 = 2.

Therefore, (1/α + 1/β) = (α + β)/(α × β) = (-6)/2 = -3.

So, the value of (1/α + 1/β) is -3.

Answer:

If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = -3

Similar Questions:

If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = ?

If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = ?

If α,β are the Zeros of the Polynomial x^2+6x+2 then (1/α+1/β) = -3

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