<s>[INST] $(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions iff $7a^2 - 50a + 7\leq 0 $ How can we show that $(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions if and only if $7a^2 - 50a + 7\leq 0$?
I know these are quadratics and can solve them, but I'm not entirely sure what the question is asking of me and how to lay out the logic.
 [/INST] From the quadratic formula you know that the solutions of
$$(a-1)x^2+3(a+1)x+4(a-1)=0$$
are given by
$$x=\frac{-3(a+1)\pm\sqrt{9(a+1)^2-16(a-1)^2}}{2(a-1)}\;.$$
These will be real if and only if
$$9(a+1)^2-16(a-1)^2\ge 0\;.$$
Expanding the lefthand side, we see that this inequality reduces to 
$$-7a^2+50a-7\ge 0\;.$$
Now just multiply the inequality by $-1$.
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