Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

up vote 5 down vote accepted

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 $-1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.