Question

If the zeroes of the quadratic polynomial p(x)= 3x + (2k-1)x-5 are equal in magnitude but opposite in sign then find the value of k

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Given: We have given that the polynomial is $p(x)= 3x^{2} + (2k-1)x-5$.

To find:We have to find the value of k in the given polynomial $p(x)= 3x^{2} + (2k-1)x-5$.

Explanation.

Step 1:As we know that the given quadratic polynomial is in the form of $ax^{2} +bx+c$ ,where $a = 3,b = (2k-1)$ and $c = -5$ respectively.

Step 2:As the zeros of the given quadratic polynomial are equal in magnitude and opposite in sign.So we have,

                   Sum of zeros $\alpha +\beta$$= 0$.

Step 3:As we know that the sum of zeros of polynomial can be given by,

⇒                                      $\alpha +\beta =\frac{-b}{a}$ $=$ coefficient of $x$/coefficient of $x^{2}$

⇒                                      $\alpha +\beta = 0$

⇒                                           $\frac{-b}{a}= 0$      

⇒So,                             $-\frac{(2k-1)}{3} = 0$  

⇒                               $-(2k-1) = 0$ × $3$  

⇒                                  $-2k+1= 0$

⇒                                        $-2k= -1$

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

⇒                                              $k= \frac{1}{2}$

∴ The value of k in the given quadratic polynomial  $p(x)= 3x^{2} + (2k-1)x-5$ is $\frac{1}{2}$.

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