Question

find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients x²-2root x-6​

Answer 1

Answer:

Hi there !!

Hi there !!p(x) = x² + 2√2x - 6

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2]

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2a = 1

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2a = 1b = 2√2

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2a = 1b = 2√2c = -6

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2a = 1b = 2√2c = -6Sum of zeros = √2 + -3√2 = -2√2 = -b/a [ -b/a = -2√2 ]

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2a = 1b = 2√2c = -6Sum of zeros = √2 + -3√2 = -2√2 = -b/a [ -b/a = -2√2 ]Product of zeros = √2 × -3√2 = - 6 = c/a [c/a = -6/1 = -6]

Hi there !!p(x) = x² + 2√2x - 6We can find the zeros by factorization [ by splitting the middle terms ]x² + 2√2x - 6x² + 3√2x - √2x - 6x [ x + 3√2 ] - √2 [ x+ 3√2][ x - √2 ] [ x+ 3√2]the zeros are = √2 and -3√2α = √2β = -3√2a = 1b = 2√2c = -6Sum of zeros = √2 + -3√2 = -2√2 = -b/a [ -b/a = -2√2 ]Product of zeros = √2 × -3√2 = - 6 = c/a [c/a = -6/1 = -6]Hope this helps you !!