If one of root of ax²+bx+c to be four times
the others prove that, 4b²= 25ac
Answers
Answer:
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.
(ii) One root is m times the other. We can assume them as t and mt.
Sum of roots = -b/a
=> t + mt = -b/a
=> (m+1)t = -b/a
=> t = -b/a(m+1) (equation 2)
Product of roots = c/a
=> t * mt = c/a
=> mt^2 = c/a
Substituting the value of t from equation 2, we'll get
=> m(-b/a(m+1))^2 = c/a
=> (mb^2)/a^2(m+1)^2 = c/a
=> (mb^2)/a(m+1)^2 = c
=> mb^2 = ac (m+1)^2
Hence, if mb^2 = ac (m+1)^2 is true, one root of ax^2 + bx + c will be m times the other.
(iii) One root is more than the other by k, which means the difference of roots will be k. We know that -
Difference of roots = √D /a
where D is discriminant of the equation, which is equal to (b^2 - 4ac). We are given that the difference of roots is k. So,
=> k = √D /a
=> k = √(b^2 - 4ac) / a
=> ak = √(b^2 - 4ac)
Squaring both the sides, we'll get
=> (ak)^2 = b^2 - 4ac
=> a^2.k^2 = b^2 - 4ac
Hence, if a^2.k^2 = b^2 - 4ac is true, one root of ax^2 + bx + c will be more than the other by k.