Question

12. If one root of equation x2 - (2k-1)x - 3 = 0 is numerically equal to the other root but opposite
in sign, then value of k is......
​

Answer 1

Answer:

Then

$\alpha = - \beta$

Sum of zeroes=

$\alpha + \beta = - \frac{b}{a}$

Thus

$\beta - \beta = 2k - 1$

$0 = 2k - 1 \\ 1 = 2k \\ k = \frac{1}{2}$

I hope it will help you.

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Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Given:-

• Equation: x²-(2k-1)x-3=0
• Both the roots are numerically equal
• But the roots have opposite sign

To Find:-

• The value of k

Solution:-

We know that the sum of zeros is (-b/a).

Here, let us take one of the zeroes as N.

It is obvious that the other zero will be (-N).

Now, the some of the the zeroes will be:-

➜ N + ( - N ) = (-(2k-1))/1

➜ 0 = -2k + 1

➜ -2k = -1

➜ k = 1/2

$\rule{200}4$