Question

If α and β are the zeroes of the polynomial x2 – 6x + k then find the value of k such that α2 + β2 = 40.​

Answer 1

f(x)=x2-6x+k;

by comparing this eq. with ax2+bx+c = 0;

we get a = 1, b = -6  and c = k;

Suppose the roots of the equation is alpha and beta:

alpha + beta = -b/a

alpha+beta = 6/1;

alpha*beta = c/a;

alpha*beta = k

we know that:

(a+b)2 = a2+b2+2ab ;

Not putting the values in above eq.;

(6)2 = 40 + 2k;

36 - 40 = 2k;

-4 = 2k;

k= -2

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Answer 2

Answer:

- 2

Step-by-step explanation:

Given :

p ( x ) = x² - 6 x + k

α² + β² = 40

We're asked to find value of 'k' :

We know :

α + β = - b  / a

= > α + β = 6

Also , α β = c / a

= > α β = k

Now :

α² + β² = ( α + β )² - 2 α β

Putting values here we get :

= > 40 = ( 6 )² - 2 × k

= > 40 = 36 - 2 k

= > - 2 k = 4

= > k = - 2

Therefore , value of k is - 2 .