what is the answer of thi question???
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p(x) = x^2 + 2x - 15
= x^2 + 5x - 3x -15
= x(x+5)-3(x+5)
= (x - 3)(x + 5)
x = 3 and x = -5
sum = -b/a
= -2
product = c/a
= -15
sum of roots = -5 + 3
= -2
product of roots = (-5) * 3
= -15
= x^2 + 5x - 3x -15
= x(x+5)-3(x+5)
= (x - 3)(x + 5)
x = 3 and x = -5
sum = -b/a
= -2
product = c/a
= -15
sum of roots = -5 + 3
= -2
product of roots = (-5) * 3
= -15
Answered by
1
x^2 + 2x - 15 = 0
x^2 + 5x - 3x - 15 = 0
x [ x + 5 ] - 3 [ x + 5 ] = 0
So,
x = -5 & x = 3
So,let the zeroes of this quadratic polynomial be α=-5 & β=3
Now,
Sum of zeroes = α+β = -b / a.
And,
Product of the zeroes = αβ = c/a
Now,
here a = 1 , b = 2 and c = -15.
Now, to verify
LHS = α+β = -5 + 3 = -2
RHS = -b/a = -[2] / 1= -2.
∴ LHS = RHS
Hence verified for the sum.
Now, for the products.
LHS =αβ= -5 * 3 = -15
RHS = c /a = -15 / 1 = -15
∴ LHS = RHS
Hence, verified for the products.
x^2 + 5x - 3x - 15 = 0
x [ x + 5 ] - 3 [ x + 5 ] = 0
So,
x = -5 & x = 3
So,let the zeroes of this quadratic polynomial be α=-5 & β=3
Now,
Sum of zeroes = α+β = -b / a.
And,
Product of the zeroes = αβ = c/a
Now,
here a = 1 , b = 2 and c = -15.
Now, to verify
LHS = α+β = -5 + 3 = -2
RHS = -b/a = -[2] / 1= -2.
∴ LHS = RHS
Hence verified for the sum.
Now, for the products.
LHS =αβ= -5 * 3 = -15
RHS = c /a = -15 / 1 = -15
∴ LHS = RHS
Hence, verified for the products.
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