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What is minimum positive value of p so that the quadratic polynomial

\[\dfrac34 x^2 - (2p+2)x  + p^2 +2 = 0\]

has real roots?

 Jun 7, 2021
 #1
avatar+208 
+1

Try and use the quadratic formula. Remember that if: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) has real roots, which part of the formula must be positive? Then substitute all the terms in.  

 Jun 7, 2021
edited by JKP1234567890  Jun 7, 2021
 #2
avatar+121054 
+2

To  have  real roots, the discriminant must be  ≥  0

 

So

 

(2p+2)^2   - 4 (3/4) ( p^2 + 2)   ≥  0

 

4p^2  + 8p  + 4  -  3p^2  - 6 ≥  0

 

p^2  +  8p  - 2  ≥  0       complete  the square on p

 

p^2  + 8p   + 16 ≥   2  + 16

 

(p + 4)^2  ≥  18       take the positive root

 

p + 4 ≥  sqrt (18)

 

p ≥  sqrt (18)  -  4   ≈   .2426 =  the  smallest  approx.  +  value of p that will give real roots

 

 

cool cool cool

 Jun 7, 2021
edited by CPhill  Jun 7, 2021
edited by CPhill  Jun 7, 2021

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