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Given the quadratic equation x2 - (A - 3)x - (A - 7), for what value of A will the sum of the squares of the roots be zero?
  • a)
    - 2
  • b)
    3
  • c)
    6
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Given the quadratic equation x2- (A - 3)x - (A - 7), for what value of...
Let the roots be p and q.
The given quadratic equation can be written as ax2 + bx + c, where a = 1, b = - (A - 3), c = - (A - 7).
The sum of the roots is (p + q) = (-b/a) = (A - 3), and the product of the roots is (pq) = (c/a) = (- A + 7).
The sum of the squares of the roots is [(p + q)2 - 2pq ] = (A - 3)2 - 2(-A + 7) = 0.
Solving this quadratic, we get A = 5 or A = -1.
Neither of these values is among the first three choices.
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Community Answer
Given the quadratic equation x2- (A - 3)x - (A - 7), for what value of...
The given quadratic equation is x^2 - (A - 3)x - (A - 7) = 0.

To find the sum of the squares of the roots, we need to find the roots of the quadratic equation and then calculate their squares.

Let's solve the quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).

Comparing the given equation with the standard form ax^2 + bx + c = 0, we have:
a = 1
b = -(A - 3)
c = -(A - 7)

Now, substituting the values of a, b, and c into the quadratic formula, we get:
x = (A - 3 ± √((A - 3)^2 - 4(1)(-(A - 7)))) / (2(1))
x = (A - 3 ± √(A^2 - 6A + 9 + 4A - 28)) / 2
x = (A - 3 ± √(A^2 - 2A - 19)) / 2

The roots of the quadratic equation are given by x1 and x2, where:
x1 = (A - 3 + √(A^2 - 2A - 19)) / 2
x2 = (A - 3 - √(A^2 - 2A - 19)) / 2

To find the sum of the squares of the roots, we calculate (x1)^2 + (x2)^2.
(x1)^2 + (x2)^2 = [(A - 3 + √(A^2 - 2A - 19)) / 2]^2 + [(A - 3 - √(A^2 - 2A - 19)) / 2]^2

Expanding and simplifying the above expression, we get:
(x1)^2 + (x2)^2 = (A^2 - 6A + 9 + 2(A - 3)√(A^2 - 2A - 19) + A^2 - 6A + 9 + 2(A - 3)√(A^2 - 2A - 19)) / 4
(x1)^2 + (x2)^2 = (2A^2 - 12A + 18 + 4(A - 3)√(A^2 - 2A - 19)) / 4
(x1)^2 + (x2)^2 = (A^2 - 6A + 9 + 2(A - 3)√(A^2 - 2A - 19)) / 2

For the sum of the squares of the roots to be zero, we need:
(A^2 - 6A + 9 + 2(A - 3)√(A^2 - 2A - 19)) / 2 = 0
A^2 - 6A + 9 + 2(A - 3)√(A^2 - 2A - 19) = 0

Simplifying the above equation:
A^2 - 6A +
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Given the quadratic equation x2- (A - 3)x - (A - 7), for what value of A will the sum of the squares of the roots be zero?a)- 2b)3c)6d)None of theseCorrect answer is option 'D'. Can you explain this answer?
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