Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        For the equation 3x^2+px+3=0,p>0,if one root is square of the other,then p is equal to?`
3 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
```
3 years ago
```							Let the roots be x and x2. By product of rootsx3 = 1x3 – 1 = 0(x – 1)(x2  - x – 1) = 0which implies x = 1 or x2 – x = -1If 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/3which implies p = 3
```
3 years ago
```							 Let the roots be x and x2. By product of rootsx3 = 1x3 – 1 = 0(x – 1)(x²+x - 1) = 0which implies x = 1 or x2 + x = -1If 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/3which implies p = 3
```
one year ago
```							Let the roots be y and y2 By the products of rootsY3=1Y3-1=0(Y-1)(Y2+Y+1) =0Which implies Y=1 or Y2+Y=-1We 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=-1We get -p/3=-1       p=3
```
one year ago
```							Discriminant>0(-p)² - 4×3×3 > 0p² - 36 > 0p² > 36p > +-6Hence p  6But given p > 0 so p > 6
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions