Question

Please solve ques no. 8th and 13th.
Please solve it a piece of paper.

From Chapter -2 Polynomial

Class 10th

I will mark as a branliest one.

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Answer 1

Heya !!!


Ques 8:-


Solution :-


P(X) => 2X²-3X+P




Here,


A = 2 , B = -3 and C = P



It is given that 3 is one zero or the Quadratic polynomial 2X²-3X+P


Let Alpha = 3


Therefore,


Sum of zeroes = -B/A


Alpha + Beta = -(-3)/2


Alpha + Beta = 3/2


3 + Beta = 3/2

Beta. = 3/2 - 3


Beta = 3-6/2

Beta = -3/2


Product of zeroes = C/A


Alpha × Beta = P/2



3 × -3/2 = P/2


-9/2 = P/2


2P = -9 × 2


2P = -18

P = -18/2 => -9



Question 13th :-


Solution:-


P(X) => X²-(K+6)X +2(2K-1)


Here,



A = 1 , B = -K-6 and C = 2(2K-1)



Sum of zeroes = -B/A



Alpha + Beta = -(-K-6)/1



Alpha + Beta = K +6 --------(1)



And,


Product of zeroes = C/A


Alpha × Beta = 2(2K-1)/1


Alpha × Beta = 4K-2 ---------(2)


According to question,



Alpha + Beta = 1/2 × (Alpha × Beta)


K +6 = 1/2 × (4K-2)



K +6 = 4K-2/2


2(K+6) = 4K-2



2K +12 = 4K -2


4K -2K = 12+2


2K = 14


K = 14/2 => 7



HOPE IT WILL HELP YOU,...... :-)

Answer 2

Heya,

here is your answer,

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8.) 2x2 - 3x + p = 0

One root is 3, so substitute x = 3
2 (3)2 - 3(3) + p = 0
18 - 9 + p = 0
9 + p = 0
p = -9

So the equation is 2x2 - 3x + p = 0
2x2 - 3x - 9 = 0

Factorising this, we get
2x2 - 6x + 3x - 9 = 0
2x ( x - 3) + 3 (x - 3) = 0
(2x + 3) (x - 3) = 0

Therefore x = -3 / 2 and 3

We know that one root is 3, therefore the other root is (-3/2)

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13.)

( Refer the pic ) :))

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Hope it helps u !!!

Cheers ☺☺

# Nikky