Question

find the zeroes of the following quadratic polynomials and verify the relationships between the zeroes and the coefficient.
${4u}^{2} + 8u$
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Answer 1

Answer:

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Step-by-step explanation:

4u2 + 8u = 4u2 + 8u + 0 = 4u(u + 2) The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2 Therefore, the zeroes of 4u2 + 8u are 0 and - 2. Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2 Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of  u2.                            

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Answer 2

Step-by-step explanation:

• relationship is sum of zero is -b/a

in this equation, -b/a( alpha+ beta) is -2

• Relationship of product of zeros is c/a(alpha X beta

in this polynomial c/a is 0

• By this we can write 2 equation

alpha+ beta=-2

alpha X beta=0

• solving this equation by substitution method we get alpha is 0& beta is -2
• Thus alpha and beta are the zeros

Hence Verified

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