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Product of real roots of the equation t2x2+|x|+9=0 [2002]
- a)is always positive
- b)is always negative
- c)does not exist
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Product of real roots of the equation t2x2+|x|+9=0 [2002]a)is always p...
Product of real roots = 
∴ Product of real roots is always positive.
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Product of real roots of the equation t2x2+|x|+9=0 [2002]a)is always p...
**Answer:**
To solve the equation t^2 * x^2 * |x| - 9 = 0, we can break it down into two cases: when x is positive and when x is negative.
**Case 1: x is positive**
In this case, the equation becomes t^2 * x^3 - 9 = 0. Solving for x, we have:
x^3 = 9/t^2
Taking the cube root of both sides, we get:
x = (9/t^2)^(1/3)
Since t^2 is always positive, the cube root of (9/t^2) will also be positive. Therefore, the product of the real roots in this case will be positive.
**Case 2: x is negative**
In this case, the equation becomes t^2 * x^3 + 9 = 0. Solving for x, we have:
x^3 = -9/t^2
Taking the cube root of both sides, we get:
x = -(9/t^2)^(1/3)
Again, since t^2 is always positive, the cube root of (-9/t^2) will also be positive. However, the negative sign in front of the cube root makes x negative. Therefore, the product of the real roots in this case will be negative.
Since the product of the real roots is positive in both cases, we can conclude that the product of the real roots is always positive. Therefore, the correct answer is option 'A'.
To solve the equation t^2 * x^2 * |x| - 9 = 0, we can break it down into two cases: when x is positive and when x is negative.
**Case 1: x is positive**
In this case, the equation becomes t^2 * x^3 - 9 = 0. Solving for x, we have:
x^3 = 9/t^2
Taking the cube root of both sides, we get:
x = (9/t^2)^(1/3)
Since t^2 is always positive, the cube root of (9/t^2) will also be positive. Therefore, the product of the real roots in this case will be positive.
**Case 2: x is negative**
In this case, the equation becomes t^2 * x^3 + 9 = 0. Solving for x, we have:
x^3 = -9/t^2
Taking the cube root of both sides, we get:
x = -(9/t^2)^(1/3)
Again, since t^2 is always positive, the cube root of (-9/t^2) will also be positive. However, the negative sign in front of the cube root makes x negative. Therefore, the product of the real roots in this case will be negative.
Since the product of the real roots is positive in both cases, we can conclude that the product of the real roots is always positive. Therefore, the correct answer is option 'A'.
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Product of real roots of the equation t2x2+|x|+9=0 [2002]a)is always positiveb)is always negativec)does not existd)none of theseCorrect answer is option 'A'. Can you explain this answer?
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