Question

If sum of the squares of zeroes of the polynomial f(t) = t^2–8t+p= 40, find the value of p.

Answer 1

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$\huge\bold\red{Answer}$

Let x and y be the zeroes of polynomial.

Then,

$\underbrace{sum\:of\:zeroes=}\underbrace\frac{-b}{a}$

$\underbrace{product\:of\:zeroes=}\underbrace\frac{c}{a}$

$compring \: with \: ax + b {x}^{2} + c$

a = 1, b = -8, c = p

hence,

x + y = 8

xy = p

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${x}^{2} + {y}^{2} = 40 \\ {(x + y)}^{2} - 2xy = 40 \\64 - 2p = 40 \\ 2p = 24 \\ p = 12$

hence,

Value of p is $\bold\green{12}$.

Answer 2

Question:

If sum of the squares of zeroes of the polynomial f(t) = t² - 8t + p is 40, find the value of  'p'.

Step-by-step explanation:

On comparing F(t) with the standard form of quadratic polynomial ax²+bx+c

we will get,

a = 1 ; b = - 8 ; c = p

Let,

α  and β  are two zeroes of given polynomial

so,

α + β = - b / a = -(-8)/1 = 8  ____ equation (1)

αβ = c /a = p / 1 = p  _____equation (2)

we have given that,

α² + β² = 40

( since,

( x + y )²=x² + y² + 2 x y

therefore,

x² + y² = ( x + y )² - 2 x y )

so,

α² + β² = 40

( α + β )² - 2 α β = 40

( by equation (1) and (2) )

(8)² - 2 ( p ) = 40

64 - 2 p = 40

- 2 p = 40 - 64

- 2 p = - 24

2 p = 24

p = 24 / 2

p = 12

HENCE THE VALUE OF 'p' IS 12.