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The equation (x2 + 6x + 7)2 + 6 (x2 + 6x + 7) + 7 = x has
- a)no real solutions
- b)exactly two real and equal solutions
- c)exactly two distinct real solutions
- d)all its solutions are real
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The equation (x2 + 6x + 7)2 + 6 (x2 + 6x + 7) + 7 = x hasa)no real sol...
To solve the equation:
(x^2 - 6x + 7)^2 - 6(x^2 - 6x + 7) - 7 = x
Let's simplify the equation step by step.
Step 1: Expand the squared term
(x^2 - 6x + 7) * (x^2 - 6x + 7) - 6(x^2 - 6x + 7) - 7 = x
(x^4 - 12x^3 + 49x^2 - 84x + 49) - 6(x^2 - 6x + 7) - 7 = x
x^4 - 12x^3 + 49x^2 - 84x + 49 - 6x^2 + 36x - 42 - 7 = x
x^4 - 12x^3 + 43x^2 - 48x = x
Step 2: Combine like terms on both sides
x^4 - 12x^3 + 43x^2 - 48x - x = 0
x^4 - 12x^3 + 43x^2 - 49x = 0
Step 3: Factor out an x
x(x^3 - 12x^2 + 43x - 49) = 0
Step 4: Solve for x using the factor theorem
x = 0 or (x^3 - 12x^2 + 43x - 49) = 0
The equation x = 0 has one real solution, which is x = 0.
Now, let's analyze the cubic equation (x^3 - 12x^2 + 43x - 49) = 0.
Using synthetic division or long division, we find that the cubic equation can be factored as:
(x - 1)(x^2 - 11x + 49) = 0
The quadratic equation x^2 - 11x + 49 does not have real solutions because its discriminant (b^2 - 4ac) is negative.
Therefore, the cubic equation has one real solution, which is x = 1.
Combining the solutions from both equations, we have x = 0 and x = 1.
Hence, the given equation has all its solutions as real numbers. Therefore, the correct answer is option D) all its solutions are real.
(x^2 - 6x + 7)^2 - 6(x^2 - 6x + 7) - 7 = x
Let's simplify the equation step by step.
Step 1: Expand the squared term
(x^2 - 6x + 7) * (x^2 - 6x + 7) - 6(x^2 - 6x + 7) - 7 = x
(x^4 - 12x^3 + 49x^2 - 84x + 49) - 6(x^2 - 6x + 7) - 7 = x
x^4 - 12x^3 + 49x^2 - 84x + 49 - 6x^2 + 36x - 42 - 7 = x
x^4 - 12x^3 + 43x^2 - 48x = x
Step 2: Combine like terms on both sides
x^4 - 12x^3 + 43x^2 - 48x - x = 0
x^4 - 12x^3 + 43x^2 - 49x = 0
Step 3: Factor out an x
x(x^3 - 12x^2 + 43x - 49) = 0
Step 4: Solve for x using the factor theorem
x = 0 or (x^3 - 12x^2 + 43x - 49) = 0
The equation x = 0 has one real solution, which is x = 0.
Now, let's analyze the cubic equation (x^3 - 12x^2 + 43x - 49) = 0.
Using synthetic division or long division, we find that the cubic equation can be factored as:
(x - 1)(x^2 - 11x + 49) = 0
The quadratic equation x^2 - 11x + 49 does not have real solutions because its discriminant (b^2 - 4ac) is negative.
Therefore, the cubic equation has one real solution, which is x = 1.
Combining the solutions from both equations, we have x = 0 and x = 1.
Hence, the given equation has all its solutions as real numbers. Therefore, the correct answer is option D) all its solutions are real.
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The equation (x2 + 6x + 7)2 + 6 (x2 + 6x + 7) + 7 = x hasa)no real solutionsb)exactly two real and equal solutionsc)exactly two distinct real solutionsd)all its solutions are realCorrect answer is option 'D'. Can you explain this answer?
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