The product of all real roots of the equation x2- |x| - 6 = 0 isa)-9b)...
Equation is x2 - |x| - 6 = 0
Case I: x > 0. then we have
x2 - x - 6 = 0 (lx| = x)
⇒ (x-3)(x+2) = 0
⇒ x = 3 is the solution as x > 0.
(So x = -2 can’t be solution)
Case II: x < 0. then we have
x2 + x - 6 = 0 (|x| = -x)
⇒ (x + 3)(x - 2) = 0
⇒ x = -3 is the solution as x < 0
(So x = 2 can't be solution)
∴ product of roots = 3. - 3 = -9
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The product of all real roots of the equation x2- |x| - 6 = 0 isa)-9b)...
To find the product of all real roots of the equation x^2 - |x| - 6 = 0, we need to first find the roots of the equation and then find their product. Let's solve the equation step by step:
1. Split the equation based on the absolute value function:
x^2 - |x| - 6 = 0
Since the absolute value of x can be either positive or negative, we have two cases to consider:
Case 1: x ≥ 0
In this case, the absolute value of x is equal to x.
x^2 - x - 6 = 0
Case 2: x < />
In this case, the absolute value of x is equal to -x.
x^2 + x - 6 = 0
2. Solve each case separately:
Case 1: x ≥ 0
To solve x^2 - x - 6 = 0, we can factorize the quadratic equation:
(x - 3)(x + 2) = 0
Setting each factor to zero, we get two possible solutions:
x - 3 = 0 -> x = 3
x + 2 = 0 -> x = -2
Case 2: x < />
To solve x^2 + x - 6 = 0, we can again factorize the quadratic equation:
(x - 2)(x + 3) = 0
Setting each factor to zero, we get two more possible solutions:
x - 2 = 0 -> x = 2
x + 3 = 0 -> x = -3
3. Combine the solutions from both cases:
The common solutions to both cases are x = -2 and x = 2.
4. Find the product of the real roots:
The product of all real roots is (-2) * (2) = -4.
Hence, the correct answer is option A) -4.