Question

If one of the zeros of the polynomial x2-px-15=30 is 3 find other root

Answer 1

Given, One root is 3
So, x = 3

Substitute x in the given equation,



x^2 - px - 15 = 30

( 3 )^2 - p( 3 ) - 15 = 30

9 - 3p - 15 = 30

9 - 15 - 30 = 3p

- 36 = 3p

- 1 2 = p






Hence,

x^2 - px - 15 - 30 = x^2 - ( - 12 × x ) - 15 - 30



=> x^2 + 12x - 45 = 0

=> x^2 + ( 15 - 3 )x - 45 = 0

=> x^2 + 15x - 3x - 45 = 0

=> x( x + 15 ) - 3( x + 15 ) = 0

=> ( x + 15 )( x - 3 ) = 0

=> x = - 15 or 3









$\boxed{ \bold{ \underline{Hence, other \: \: root \: \: is \: \: - 15}}}$

Answer 2

The given equation is : X² - PX - 15 = 30




P ( X ) = X² - PX - 45 = 0



Substitute X = 3 in P ( x ).


P (3) = (3)² - P × 3 - 45 = 0



=> 9 - 3P - 45 = 0



=> -3P - 36 = 0



=> -3P = 36


=> P = -12



P ( X ) = X² - (-12) × X - 45


P ( X ) = X² + 12X - 45



Sum of zeroes = -B/A



3 + Other zero = -12


Other zero = -15