find a quadratic polynomial each with the given number as the sum and the product of it’s zeros respectively (i) 4 and -2 (ii) 0 a

find a quadratic polynomial each with the given number as the sum and the product of it’s zeros respectively (i) 4 and -2 (ii) 0 and -10/3 (iii) 5/7 and 0 (iv) -5 and -6 (v) √2 and -12 (vi) 3 and -2 (vii) -2√3 and -9 (viii) -3/2√5 and 1/2​

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  1. Given

    we have given sum and Product of a quadratic polynomial:(i) 4 and -2 (ii) 0 and -10/3 (iii) 5/7 and 0 (iv) -5 and -6 (v) √2 and -12 (vi) 3 and -2 (vii) -2√3 and -9 (viii) -3/2√5 and 1/2

    To Find

    we have to find the quadratic polynomial

    [tex]\sf\huge {\underline{\underline{{Solution}}}}[/tex]

    Since, we have given sum and Product of roots of the quadratic polynomial so, we can use this formula to find out the Equation.

    x²-(sum of roots)x+ (product of roots)=0

    (I) 4 and -2

    x²-(4)x+(-2)=0

    => x²-4x-2=0

    (ii) 0 and -10/3

    x²-(0)x+(-10/3)=0

    => x²-10/3=0

    (iii) 5/7 and 0

    x²-(5/7)x+(0) =0

    => x²-5/7x=0

    (iv) -5 and -6

    x²-(-5)x+(-6)=0

    => x²+5x-6=0

    (v) √2 and -12

    x²-(√2)x+(-12)=0

    =>x²-√2x-12=0

    (vi) 3 and -2

    x²-(3)x+(-2)=0

    =>x²-3x-2=0

    (vii) -2√3 and -9

    x²-(-2√3)x+(-9)=0

    =>x²+2√3x-9=0

    (viii)-3/2√5 and 1/2

    x²-(-3/2√5)x+(1/2)=0

    =>x²+3/2√5x+1/2=0

    Check:

    we can check by factorise the obtained polynomial

    (iv) –5 and -6

    x²-(-5)x+(-6)=0

    => x²+5x-6=0

    Now factorise it

    x= -b±√ b²-4ac/2a

    a= 1 ,b= 5 & c = -6

    x= -5±√5²-4(1)(-6)/2

    x= -5±√ 25-(-24)/2

    x= -5±√ 25+24/2

    x= -5±√49/2

    x= -5±7/2

    x= -5+7/2 or x = -5-7/2

    x= 2/2 or x= -12/2

    x= 1 or x = -6

    sum of zeroes = 6+1= 5

    product of zeroes= 6*1= 6

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