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dcfcviper · Nov 1, 2007, 08:22 AM

Stuck again on something different this time :(

Find all values of b for which the equation $4x^2 - bx + 13 = 0$ has 2 real roots.

I tried the quadratic formula and got $x=\frac{b\pm\sqrt{b^2-208}}{8}$ and subbed that back in but get down to $-b^2 + 4b + 3sqrt{b^2 -208} +208 = 0$ for the positive value.

ebaines · Nov 1, 2007, 09:04 AM

In order for the quadratic formula to yield two real roots, the term $sqrt {b^2 - 4ac}$ must be real and greater than 0. That means $b^2-4ac$ must be greater than 0. Hope this helps.

dcfcviper · Nov 1, 2007, 09:53 AM

So I get $b<-sqrt{208}$ or $b>sqrt{208}$

But is that right? It says there is 10 marks for the question.

ebaines · Nov 1, 2007, 09:57 AM

Looks good to me. It's always a good idea to try a few values around your answer to see if it works as you expect. For example, sqrt(208) is around 14.42, so try to see if the original quadratic equation has two real roots if b= -15, -14, +14, and +15.

dcfcviper · Nov 1, 2007, 10:26 AM

Checked it out with those values and the results fit with $b<-sqrt{208}$ and $b>sqrt{208}$ with $\pm\ 14$ giving an imaginary root and $\pm\ 15$ giving 2 real root so it works.

Thnks very much again ebaines :thup: