how do you solve sqroot(3x - 1) > x + 7?I can take it to where I have 0 > x^2 + 11x + 50 but then I can't figure out how to factor it
Already Tried: I squared both sides to get rid of the radical, and then consolidated to 0 > x^2 + 11x + 50, but now I can't figure out how to factor that to get to an Interval Notation. thanks
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√(3x - 1) > x + 7
Squre both sides
3x - 1 > (x + 7)^2
3x - 1 > x^2 + 14x + 49
0 > x^2 + 11x + 50
x^2 + 11x + 50 < 0
The left hand side is a quadratic with a = 1, b = 11, c = 50
Its discriminant is b^2 - 4ac = 11^2 - 4 * 1 * 50 = -79
Since the discriminant is < 0, the quadratic has no real solutions
This means the inequality x^2 + 11x + 50 < 0 has no solution
Answer: √(3x + 1) > x + 7 has no solution
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