Question

If one root of x^2-x=k(2x+1) is multiplicative inverse of other then find the value of k

Answer 1

Given :

• Equation is x² - x = k(2x + 1)
• One root is multiplicative inverse of other

To Find :

• Value of k

Solution :

⇒x² - x = k(2x + 1)

⇒x² - x = 2kx + k

⇒x² - x - 2kx - k = 0

⇒x² - (2k + 1)x - k = 0 ...(1)

As we know that the general form of quadratic polynomial is ax² + bx + c

On comparing (1) with general equation,

• a = 1
• b = -(2k + 1)
• c = -k

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Let the zeroes be α and β

Then,

⇒β = 1/α ...(2)

Now, use formula for product of zeroes

$\implies \sf{Product \: of \: zeros \: = \: \dfrac{c}{a}} \\ \\ \implies \sf{\alpha \: \times \: \beta \: = \: \dfrac{-k}{1}} \\ \\ \implies \sf{\alpha \: \times \: \dfrac{1}{\alpha} \: = \: -k} \\ \\ \implies \sf{-1 \: = \: k}$

$\therefore$ Value of k is -1.

Answer 2

Given: One root of the equation x² - x = k(2x + 1) is the multiplicative inverse of the other.

To find: The value of k.

Answer:

We know that the general form of an equation is ax² + bx + c, where:

• The sum of the zeros is -b/a.
• The product of the zeros is c/a.

x² - x = k(2x + 1) is the equation given to us.

Modifying it, we get x² - x - k(2x + 1), where:

• a = 1
• b = -1
• c = -k

Now, we know that one of the zeros is the inverse of the other.

Let's assume that one of the zeros is α. The other one will be 1/α as per the question.

Equating it to the product of the zeros:

$\tt \implies\ \alpha\ \times\ \dfrac{1}{\alpha}\ =\ \dfrac{-k}{1}\\\\\\1\ =\ \dfrac{-k}{1}\\\\\\1\ =\ -k\\\\\\\bf -1\ =\ k$

Therefore, k = -1.