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Question

If the sum of the squares of zeros of quadratic polynomial F (X)= x^2 - 8 x + K is 40, find k

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Solution

p (x)= x^2-8x+k
p (x)=Ax^2+Bx+C(The equation is in this form)

Let the zeroes be 'a' and 'b'

It is given that
=> a^2+b^2=40
=> (a+b)^2 - 2ab = 40 ----1

sum of zeroes
=> a+b = -B/A = -(-8)/1 = 8

Product of zeroes
=> a*b = C/A = k

Substitute these values in the equation1 we get
=> 8^2 - 2k = 40
=> 64 - 2k = 40
=> 2k = 24
=> k = 12

therefore the value of k is 12




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