Question

find the zeros of the quadratic polynomial x^2+x-12 and verify the relation between zeros and coefficient
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Answer 1

Answer:

→ x^2 + x - 12

→ x^2 + 4x - 3x - 12

→ x( x + 4 ) - 3( x + 4 )

→ ( x + 4 ) ( x - 3 )

Hence roots of this polynomial are - 4 and 3.

Sum of roots = -b/a { for standard eq. }

→ - 4 + 3 = -1/1

→ - 1 = - 1, which is true.

Product of roots = c/a

→ - 4 * 3 = - 12 / 1

→ - 12 = - 12, which is true.

Hence roots are - 4 and 3 and relation is now verified as well.

Answer 2

Answer:

heyyy,

we have to find the zeroes of the quadratic polynomial, x² + x - 12.

x² + x - 12 = 0

⇒x² + 4x - 3x - 12 = 0

⇒x(x + 4) - 3(x + 4) = 0

⇒(x - 3)(x + 4) = 0

⇒x = 3, -4

sum of zeroes = - coefficient of x/coefficient of x²

sum of zeroes = 3 - 4 = -1

- coefficient of x/coefficient of x² = -1/1 = -1

LHS = RHS

product of zeroes = constant/coefficient of x²

product of zeros = 3 × -4 = -12

constant/coefficient of x² = -12/1 = -12

LHS = RHS

hence verified

Step-by-step explanation:

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