Question

Find the value of q so that equation 2xsquare-3px+5q=0 has one root which is twice the other

Answer 1

Step-by-step explanation:

Given Find the value of q so that equation 2 x ^2 - 3 q x + 5 q=0 has one root which is twice the other

• We know that standard form of equation is ax^2 + bx + c = 0
• Equation is 2x^2 – 3qx + 5q = 0 ----------1
• From 1 we get
• So a = 2, b = - 3q, c = 5q
• Let m,n be the roots of the equation
• We know that m + n = - m/n and m.n = c/a
• Since, one root is twice the other n = 2m
• So m + 2m = - (- 3q) / 2 and m.2m = c/a
•   3m = 3q / 2 and 2m^2 = 5q / 2
• So m = q/2 and 4m^2 – 5q = 0
• Now 4 (q^2 / 4) – 5q = 0
• So q^2 – 5q = 0
• Or q(q – 5) = 0
• Or q = 5

Reference link will be

https://brainly.in/question/15294382