Discriminant question help?

[NewAccount]

2qx^2 + 5qx + 5q - 3 = 0 has two real roots, find the range of values of q

I found that q is less than 8/5 but I don’t k ow how to prove that q is greater than 0

 

[Deleted member 4993]

2qx^2 + 5qx + 5q - 3 = 0 has two real roots, find the range of values of q

I found that q is less than 8/5 but I don’t k ow how to prove that q is greater than 0

How did you find "q is less than 8/5"?

Please share your work.

 

[pka]

2qx^2 + 5qx + 5q - 3 = 0 has two real roots, find the range of values of q
I found that q is less than 8/5 but I don’t k ow how to prove that q is greater than 0

Have a look at this link.

 

[Dr.Peterson]

I strongly suspect that when you tried solving a quadratic inequality, you just divided by q, rather than using the method you were taught, which would involve factoring. Dividing by q without changing the direction of the inequality assumes that q > 0!

 

[HallsofIvy]

By the "quadratic formula" the roots of $\displaystyle 2qx^2 + 5qx + 5q - 3 = 0$ are given by $\displaystyle x= \frac{-5q\pm\sqrt{25q^2- 8q(5q- 3)}}{4q}$. The roots are real as long and distinct as the "discriminant" is strictly positive: $\displaystyle 25q^2- 8q(5q- 3)= 24q- 15q^2= 3q(8- 5q) \ge 0$. The product of two numbers is positive if and only if the two numbers have the same sign- both positive or both negative.

Both are positive when 3q> 0 and 8- 5q> 0. From the second, 8>5q so q< 8/5. From the first, q> 0 so 0< q< 8/5.

Both are negative when 3q< 0 and 8- 5q< 0. From the second, 8< 5q so q> 8/5. From the first q< 0. It is impossible to satisfy both so this give no values of q.

0< q< 8/5.