Question

sum of squares of zeros of the quadratic polynomial p(x)=x2-8x+k is 40 find the value of k

Answer 1

Given f(x) = x^2 - 8x + k.

It is in the form of ax^2 + bx + c,

where a = 1, b = -8, c = k.

Let a and be be the zeroes of the quadratic polynomial.

Now,

Sum of roots = -b/a

a + b = -(-8)/1

a + b = 8  ---- (1)



Product of roots = c/a

ab = k  ----- (2)


Given Sum of squares is 40.

a^2 + b^2 = 40.

Therefore:

(a + b)^2 = a^2 + b^2 + 2ab

64 = 40 + 2(k)

64 - 40 = 2k

24 = 2k

k = 12.


Therefore k = 12.


Hope this helps!

Answer 2

Let the zeroes be x and y

Sum of zeroes = - b/a

x + y = - (-8) / 1

=> x + y = 8 ----(1)

Product of zeroes = c/a

=> xy = k / 1

=> xy = k ----(2)


x^2 + y^2 = 40 ( Given)

On adding ( 2xy) in both sides, we get


x^2 + y^2 + 2xy = 40 + 2xy

=> ( x + y)^2 = 40 + 2k

=> ( 8) ^2 = 40 + 2k

=> 64 = 40 + 2k

=> 24 = 2k

=> k = 12