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For the equation 3x^2+px+3=0,p>0,if one root is square of the other,then p is equal to?

For the equation 3x^2+px+3=0,p>0,if one root is square of the other,then p is equal to?

Grade:12

6 Answers

Dhruv
6 Points
7 years ago
Let two roots are x and xsquare so by product of root we can find value of x as 1,w ,wsquare And by sum of roots value of p :But we have to reject 1 as p>0 so by work wsquare we have to solveW+ Wsquare=-p/3-1=-p/3P =3
anuraag
19 Points
6 years ago
Let the roots be x and x2. 
By product of roots
x3 = 1
x3 – 1 = 0
(x – 1)(x2  - x – 1) = 0
which implies x = 1 or x2 – x = -1
If we take x = 1 then by sum of roots p = -6 (but it is rejected as p > 0)
If we take  x2 – x = -1 then by equating with sum of roots -1 = -p/3
which implies p = 3
Rajeev
13 Points
5 years ago
 
Let the roots be x and x2. 
By product of roots
x3 = 1
x3 – 1 = 0
(x – 1)(x²+x - 1) = 0
which implies x = 1 or x2 + x = -1
If we take x = 1 then by sum of roots p = -6 (but it is rejected as p > 0)
If we take  x2 +x = -1 then by equating with sum of roots -1 = -p/3
which implies p = 3
Shubham kumar
13 Points
5 years ago
Let the roots be y and y2 
By the products of roots
Y3=1
Y3-1=0
(Y-1)(Y2+Y+1) =0
Which implies Y=1 or Y2+Y=-1
We can't apply Y=1 because by applying that we get p=-6 (which is not correct p>0) 
So by equating the sum of roots and Y2+Y=-1
We get -p/3=-1       p=3  
Utsav Jha
13 Points
5 years ago
Discriminant>0
(-p)² - 4×3×3 > 0
p² - 36 > 0
p² > 36
p > +-6
Hence p 6
But given p > 0 so p > 6
 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Let α,α2 be the roots of 3x^2 + px + 3 = 0
Now, S = α + α^2 = −3p​,
P = α^3 = 1
⇒ α = 1 , ω , ω^2
Now, α + α^2 = 3p​
⇒ ω + ω^2 = −3p
​⇒ −1 = −3p ​
⇒ p = 3

Thanks and Regards

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