Question

if one root is 4 times the other root of the equation ax2+bx+c=0 then prove that 4b²=25a​

Answer 1

One root is 4 times the other. We can assume them as t and 4t.

Sum of roots = -b/a

=> t + 4t = -b/a

=> 5t = -b/a

=> t = -b/5a (equation 1)

Product of roots = c/a

=> t * 4t = c/a

=> 4t^2 = c/a

Substituting the value of t from equation 1, we'll get

=> 4(-b/5a)^2 = c/a

=> (4b^2)/(25a^2) = c/a

=> (4b^2)/25a = c

=> 4b^2 = 25ac

Hence, if 4b^2 = 25ac is true, one root of ax^2 + bx + c will be 4 times the other.

Answer 2

Refer attachment for your answer