Find all a such that the equation x^2 -2(a-2)x +3a +4= 0 has two real solutions

Question

Find all a such that the equation x^2 -2(a-2)x +3a +4= 0 has two real solutions both of which greater than 1.

Explanation / Answer

Using the quadratic equation, we know when solutions are real whenB^2 - 4AC >=0. To have two different solutions, B^2 - 4AC>0. Substituting our values, we have B^2 - 4AC = (-2(a-2))^2 - 4 * 1 *(3a +4) = 4a^2-16a+16 -12a -16 =4a^2 - 28a > 0, giving a^2 - 7a > 0, so   a^2 > 7a. If a is positive, this gives a>7. If a is negative, thisgives a2 a-2 - (a^2 - 7a) > 1 - (a^2 - 7a) > 3 - a a^2 - 7a > (3-a)^2 a^2 - 7a > 9 - 6a + a^2 - 7a > 9 - 6a -9 > a. So, since -9>a, and then either a>7 or a

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