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If x^2 + 2ax + 10 - 3a > 0 for all real numbers x, then find all possible values of a.

 Jun 14, 2020
 #1
avatar+8341 
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If \(x^2 + 2ax + 10 - 3a > 0\) for all real numbers x, then this means the graph of \(f(x) = x^2 + 2ax + 10 - 3a\) never touches the x-axis. Because if it touches the x-axis, at that point, the function would be zero. However, the questions says that f(x) > 0, so it is never zero, implying it never touches the x-axis.

 

For quadratic functions that doesn't touch the x-axis, we always have \(\Delta < 0\), where \(\Delta\) is the discriminant.

 

We calculate \(\Delta\).

 

\(\Delta = (2a)^2 - 4(1)(10 - 3a) = 4a^2 + 12a - 40 = 4(a^2 + 3a - 10)\)

 

If we solve the inequality \(\Delta < 0\), we get \(-5 < a < 2\) as our answer. (how?)

 Jun 14, 2020
 #2
avatar+1130 
+1

x=−a+√[(a−2)(a+5)]

x=−a−√[(a−2)(a+5)]

 

so x can be anything when x^(2) + 2ax- 3a >-10 is true where we substitute x with the equation above to get x=(-5)<=>(2)

 Jun 14, 2020

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