Question

plz guyes solve this two plz for 50 points plz do fast ​

Answer 1

Hey mate!

Here is yr answer....

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Let, α , β are the zeros...

(1)

α+β = -8/3

α × β = 4/3

According to quadratic formula,

K [ x² -(α+β)x + α×β ]

= K [x² -(-8/3)x + 4/3]

= K [x² + 8x/3 + 4/3]

= K [3x²+8x+4 /3}

let , K = 3

= 3 [3x²+8x+4 /3]

= 3x²+8x+4

Therefore, 3x²+8x+4 is the required quadratic polynomial

Let, 3x²+8x+4 = 0

3x²+6x+2x+4 = 0

3x(x+2) + 2(x+2) = 0

(x+2) (3x+2) = 0

x = -2 or x = -2/3

Therefore, -2 and -2/3 are the zeros..
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(iv)

α +β = -3/2√5

α ×β = -1/2

According to quadratic formula,

K [x² -(α+β)x +α×β ]

= K [x² -(-3/2√5)x + (-1/2)]

= K [x² + 3x/2√5 -1/2]

= K [2√5x²+3x-√5 /2√5]

let , K = 2√5

= 2√5 [2√5x²+3x-√5 /2√5]

= 2√5x²+3x-√5

Therefore, 2√5x²+3x-√5 is the required quadratic polynomial..

Let, 2√5x²+3x-√5 = 0

2√5x²-2x +5x -√5 = 0

2x(√5x-1) + √5(√5x-1) = 0

(2x+√5) (√5x-1) = 0

x = -√5/2 or x= 1/√5

Therefore, -√5/2 and 1/√5 are the zeros of the polynomial

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Hope it helps...

Answer 2

Answer:

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