# $(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ 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.

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The general idea is to find the set of $a$ for which the first equation has real solutions, and show that it is equal to the set of $a$ for which the inequality in the second equation is satisfied. –  Jonathan Christensen Dec 4 '12 at 1:22

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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