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The quadratic polynomial whose sum of zeroes is 3 and the product of zeroes is –2 is :
- a)x2 + 3x – 2
- b)x2 – 2x + 3
- c)x2 – 3x + 2
- d)x2 – 3x – 2
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The quadratic polynomial whose sum of zeroes is 3 and the product of z...
Sum of zeros = 3/1
-b/a = 3/1 .....................(1)
-b/a = 3/1 .....................(1)
Product of zeros = -2/1
c/a = -2/1 ...................(2)
c/a = -2/1 ...................(2)
From equation (1) and (2)
a = 1
-b = 3, b = -3
c = -2
a = 1
-b = 3, b = -3
c = -2
The required quadratic equation is
ax2+b
= x2-3x-2
ax2+b
x
+c= x2-3x-2
Most Upvoted Answer
The quadratic polynomial whose sum of zeroes is 3 and the product of z...
P(x)=x^2- (sum of zeroes)x + Product of zeroes =x^2 - (3)x + (-2) =x^2 - 3x - 2 ..... Clearly, Correct Answer is Option(d).. Option(B) is not the correct answer..
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Community Answer
The quadratic polynomial whose sum of zeroes is 3 and the product of z...
There are different ways to approach this problem, but one common method is to use the fact that a quadratic polynomial with roots x1 and x2 can be written as:
p(x) = a(x - x1)(x - x2)
where a is a non-zero constant (the leading coefficient). This expression shows that the roots of p(x) are x1 and x2, and the product of the roots is:
x1x2 = -a/c
where c is the constant term of p(x). To find the polynomial with sum of roots 3 and product of roots k, we can set up a system of equations:
x1 + x2 = 3
x1x2 = k
We can solve for x1 and x2 using the quadratic formula:
x1,2 = (3 ± √(9 - 4k))/2
Note that for real roots to exist, we must have 9 - 4k ≥ 0, or k ≤ 9/4. We can then write the polynomial as:
p(x) = a(x - x1)(x - x2)
= a(x^2 - (x1 + x2)x + x1x2)
= a(x^2 - 3x + k)
where we substitute the values of x1 and x2 in terms of k. To determine the value of a, we can use the fact that the leading coefficient is a:
a = p(1)
where we evaluate the polynomial at x = 1. Since we know the roots and the product of the roots, we can compute:
p(x) = (x - x1)(x - x2)
= x^2 - (x1 + x2)x + x1x2
= x^2 - 3x + k
and therefore:
p(1) = 1 - 3 + k = k - 2
Thus, the quadratic polynomial with sum of roots 3 and product of roots k is:
p(x) = (x^2 - 3x + k)/(k - 2)
where k is a real number such that k ≤ 9/4.
p(x) = a(x - x1)(x - x2)
where a is a non-zero constant (the leading coefficient). This expression shows that the roots of p(x) are x1 and x2, and the product of the roots is:
x1x2 = -a/c
where c is the constant term of p(x). To find the polynomial with sum of roots 3 and product of roots k, we can set up a system of equations:
x1 + x2 = 3
x1x2 = k
We can solve for x1 and x2 using the quadratic formula:
x1,2 = (3 ± √(9 - 4k))/2
Note that for real roots to exist, we must have 9 - 4k ≥ 0, or k ≤ 9/4. We can then write the polynomial as:
p(x) = a(x - x1)(x - x2)
= a(x^2 - (x1 + x2)x + x1x2)
= a(x^2 - 3x + k)
where we substitute the values of x1 and x2 in terms of k. To determine the value of a, we can use the fact that the leading coefficient is a:
a = p(1)
where we evaluate the polynomial at x = 1. Since we know the roots and the product of the roots, we can compute:
p(x) = (x - x1)(x - x2)
= x^2 - (x1 + x2)x + x1x2
= x^2 - 3x + k
and therefore:
p(1) = 1 - 3 + k = k - 2
Thus, the quadratic polynomial with sum of roots 3 and product of roots k is:
p(x) = (x^2 - 3x + k)/(k - 2)
where k is a real number such that k ≤ 9/4.
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The quadratic polynomial whose sum of zeroes is 3 and the product of zeroes is –2 is :a)x2+ 3x – 2b)x2– 2x + 3c)x2– 3x + 2d)x2– 3x – 2Correct answer is option 'D'. Can you explain this answer?
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