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  • Find the value of k such that polynomial  has sum of its zeros equal to half of their product.  

Find the value of k such that polynomial x^2-(k+6)x+2(2k-1) has sum of its zeros equal to half of their product.

 

 
 
 
 
 

Answers (1)

The standard form of a quadratic equation = ax^2+bx+c=0

Product of zeros = c/a ----(1)

Sum of zeros = -b/a ----(2)

Comparing the equations

a=1, b=(-k+6), c=2(2k-1)

After putting the values in eqn1 and eqn 2

Product of zeros = \frac{2\left ( 2k-1 \right )}{1}= 2\left ( 2k-1 \right )

Sum of zeros = \frac{-\left [ -\left ( k+6 \right ) \right ]}{1}= k+6

As per the given statement in the question 

Sum of zeros = 1/2 X Product of zeros

k+6 = \frac{1}{2}\left [ 2\left ( 2k-1 \right ) \right ]

k+6 =2k-1

k=7

Hence the value of k will be 7.

Posted by

Safeer PP

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