Question

Given the quadratic equation ax²-5b+4a=0, where a and b are constants has two real an equal roots,find a:b​

Answer 1

Therefore either a:b = 5:4 or a:b=-5:4

Explanation:

ax²-5bx+4a=0

Since the quadratic equation has two real root.

Then b²-4ac>0

Here a= a , b= -5b and c=4a

∴(-5b)²-4.a.4a=0

⇔25b²=16a²

⇔5b=±4a

`\Leftrightarrow (a)/(b) =\pm(5)/(4)`

Therefore either a:b = 5:4 or a:b=-5:4