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Find the zeroes of the quadratic polynomial and verify the relationship between zeroes and coefficients of (i): root3x^2 10x 7root3 (ii) x^2 2root2x-6 Pls someone help me out with this question because its urgent?
Most Upvoted Answer
Find the zeroes of the quadratic polynomial and verify the relationshi...
P(x) = √3x2+10x+7√3
= √3x+3x+7x+7√3
= (√3x+7)(x+√3)
= √3x+7 = 0 and x+√3 = 0
= x = -7/√3 and x = -√3
verification by α and β,
let α = -7/√3 and β = -√3
sum of zeros = α+β = -7/√3+(-√3)
= -7/√3-√3
= -7-3/√3
= -10/√3
product of zeroes = αβ
= -7/√3 . -√3
= 7
p(x) = x� + 2√2x - 6
We can find the zeros by factorization [ by splitting the middle terms ]
x� + 2√2x - 6
x� + 3√2x - √2x - 6
x [ x + 3√2 ] - √2 [ x+ 3√2]
[ x - √2 ] [ x+ 3√2]
the zeros are = √2 and -3√2
α = √2
β = -3√2
a = 1
b = 2√2
c = -6
Sum 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]
Community Answer
Find the zeroes of the quadratic polynomial and verify the relationshi...
Finding the zeroes of quadratic polynomials and verifying the relationship between zeroes and coefficients:
To find the zeroes of a quadratic polynomial, we need to set the polynomial equal to zero and solve for x. The zeroes of the polynomial are the values of x that make the polynomial equal to zero.
For the polynomial root3x^2 + 10x + 7root3:
Step 1: Set the polynomial equal to zero
root3x^2 + 10x + 7root3 = 0
Step 2: Solve for x
Using the quadratic formula, we get:
x = [-b ± sqrt(b^2 - 4ac)]/2a
Where a = root3, b = 10, and c = 7root3
x = [-10 ± sqrt(10^2 - 4(root3)(7root3))]/2(root3)
x = [-10 ± sqrt(100 - 84)]/2(root3)
x = [-10 ± sqrt(16)]/2(root3)
x = [-10 ± 4]/2(root3)
x1 = (-10 + 4)/2(root3) = -3/root3
x2 = (-10 - 4)/2(root3) = -7/root3
Therefore, the zeroes of the polynomial are -3/root3 and -7/root3.
Step 3: Verify the relationship between zeroes and coefficients
The relationship between the zeroes and coefficients of a quadratic polynomial is given by:
x1 + x2 = -b/a
x1 * x2 = c/a
Plugging in the values of a, b, and c from the given polynomial, we get:
x1 + x2 = -(10/root3)/(root3) = -10/3
x1 * x2 = (7root3)/(root3) = 7
Verifying the above relationship, we get:
(-3/root3) + (-7/root3) = -10/3
(-3/root3) * (-7/root3) = 7
Therefore, the relationship is verified.
For the polynomial x^2 + 2root2x - 6:
Step 1: Set the polynomial equal to zero
x^2 + 2root2x - 6 = 0
Step 2: Solve for x
Using the quadratic formula, we get:
x = [-b ± sqrt(b^2 - 4ac)]/2a
Where a = 1, b = 2root2, and c = -6
x = [-2root2 ± sqrt((2root2)^2 - 4(1)(-6))]/2(1)
x = [-2root2 ± sqrt(32)]/2
x = -root2 ± 2sqrt2
Therefore, the zeroes of the polynomial are -root2 + 2sqrt2 and -root2 - 2sqrt2.
Step 3: Verify the relationship between zeroes and coefficients
The relationship between the zeroes and coefficients of a quadratic polynomial is given by:
x1 + x2 = -b/a
x1 * x2 = c/a
Plugging in the values of a, b, and c from the given polynomial, we get:
x1 + x2 = -2root2/1 = -
To find the zeroes of a quadratic polynomial, we need to set the polynomial equal to zero and solve for x. The zeroes of the polynomial are the values of x that make the polynomial equal to zero.
For the polynomial root3x^2 + 10x + 7root3:
Step 1: Set the polynomial equal to zero
root3x^2 + 10x + 7root3 = 0
Step 2: Solve for x
Using the quadratic formula, we get:
x = [-b ± sqrt(b^2 - 4ac)]/2a
Where a = root3, b = 10, and c = 7root3
x = [-10 ± sqrt(10^2 - 4(root3)(7root3))]/2(root3)
x = [-10 ± sqrt(100 - 84)]/2(root3)
x = [-10 ± sqrt(16)]/2(root3)
x = [-10 ± 4]/2(root3)
x1 = (-10 + 4)/2(root3) = -3/root3
x2 = (-10 - 4)/2(root3) = -7/root3
Therefore, the zeroes of the polynomial are -3/root3 and -7/root3.
Step 3: Verify the relationship between zeroes and coefficients
The relationship between the zeroes and coefficients of a quadratic polynomial is given by:
x1 + x2 = -b/a
x1 * x2 = c/a
Plugging in the values of a, b, and c from the given polynomial, we get:
x1 + x2 = -(10/root3)/(root3) = -10/3
x1 * x2 = (7root3)/(root3) = 7
Verifying the above relationship, we get:
(-3/root3) + (-7/root3) = -10/3
(-3/root3) * (-7/root3) = 7
Therefore, the relationship is verified.
For the polynomial x^2 + 2root2x - 6:
Step 1: Set the polynomial equal to zero
x^2 + 2root2x - 6 = 0
Step 2: Solve for x
Using the quadratic formula, we get:
x = [-b ± sqrt(b^2 - 4ac)]/2a
Where a = 1, b = 2root2, and c = -6
x = [-2root2 ± sqrt((2root2)^2 - 4(1)(-6))]/2(1)
x = [-2root2 ± sqrt(32)]/2
x = -root2 ± 2sqrt2
Therefore, the zeroes of the polynomial are -root2 + 2sqrt2 and -root2 - 2sqrt2.
Step 3: Verify the relationship between zeroes and coefficients
The relationship between the zeroes and coefficients of a quadratic polynomial is given by:
x1 + x2 = -b/a
x1 * x2 = c/a
Plugging in the values of a, b, and c from the given polynomial, we get:
x1 + x2 = -2root2/1 = -
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