Question

8. If one of the zeroes of Polynomial P(x): x²-8x+kis seven times of other then find the value of K.
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Answer 1

HERE IS YOUR ANSWER MATE.....;

$eq {}^{n} = > x {}^{2} - 8x + k$

let one zero = y

other zero = 7y

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

$= > 7y + y = - \frac{ - 8}{1}$

$= > 8y = 8$

$= > y = 1$

one zero = y = 1

other zero = 7y= 7(1) = 7

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

$= > 7 \times 1 = \frac{k}{1}$

$= > k = 7$

HOPE IT'S HELPFUL....:-)

Answer 2

Answer:

HERE IS YOUR ANSWER MATE:-

$eq {}^{n} = > x {}^{2}$

let one zero = y

other zero = 7y

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

= > $7y + y = - \frac{ - 8}{1}$

= > $8y = 8=>8y=8$

= > $y = 1=>y=1$

one zero =$y = 1$

other zero = $7y= 7(1) = 7$

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

= > $7 \times 1 = \frac{k}{1}$

= > $k = 7=>k=7$

HOPE IT'S HELP YOU A LOT :)