Question

a one of the root of equation X square + PX + Q is equal to zero is three times the other prove that three p square is equal to 16 q​

Answer 1

Let roots are a and a^2. Therefore

a+a^2=-p………………..(1)

a.a^2=q

a^3=q……………………(2)

On cubing eq.(1)both side

a^3+3a.a^2.(a+a^2)+a^6=-p^3

a^3+3.a^3.(a+a^2)+(a^3)^2 =-p^3

put a^3= q , and (a+a^2)=-p

q-3.q.p+q^2=-p^3

p^3–3.q.p+q+q^2=0

p^3-q(3p-1)+q^2=0 , proved.


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Answer 2

Step-by-step explanation:

Given one of the root of equation x^2 + PX + Q is equal to zero is three times the other prove that three p square is equal to 16 q

• According to the equation x^2 + px + q = 0
• So roots are a and 3a
• So a + b = - b / a
• Or a + 3a = - p
• Or 4a = - p
• Or a = - p / 4
• Now ab = c/a
• Or a x 3a = q/1
• Or 3a^2 = q
• Or 3 (- p/4)^2 = q
• 3p^2 / 16 = q
• Or 3p^2 = 16 q

Reference link will be

https://brainly.in/question/1083385