Show that for $ 3 > y_1 >0 $ the roots of the equation
$$(y_1-2)x^2-(8-2y_1)x-(8-3y_1)=0$$ are real, where $y_1$ is a constant.
Due to my difficulties in doing this I would be grateful for your help.
Asked
Active
Viewed 47 times
0
-
1Hint: evaluate the determinant $\Delta = b^2 - 4ac.$ Which assumptions can be made? – Luigi D. Mar 12 '15 at 09:57
1 Answers
0
The question asks us to show if the roots of the equation (y1−2)x^2−(8−2y1)x−(8−3y1)=0 are real when 3>y1>0 which means:
b^2-4ac>0 or b^2-4ac=0 when y1 is any number between 0 and 3
Therefore: we can replace y1 by 0, 1, 2 or 3 and it will give an answer like zero or a number which is greater than 0.
For e.g Lets take y1= 1
Replace in the equation which will give
-x^2-6-5=0
Using b^2-4ac
= (-6)^2-4(-1)(-5)
= 36-20
= 16
Since 16>0 the roots of the equation are real.