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