Showing that 3x^2-x+k=0 has roots \frac{p}{4} and p+1 for unique

predorsomwy

predorsomwy

Answered question

2022-04-25

Showing that 3x2x+k=0 has roots p4 and p+1 for unique values of k and p

Answer & Explanation

Jayla Matthews

Jayla Matthews

Beginner2022-04-26Added 18 answers

A quadratic ax2+bx+c with roots x1,x2 factors as a(xx1)(xx2). In this case:
 3x2x+k=3(xp4)(x(p+1))      12x24x+4k=3(4xp)(x(p+1))      12x24x+4k=12x2(15p+12)x+p(p+1)      {4=15p+124k=p(p+1)
The first equation gives p=815, then the second one gives k.

Alice Harmon

Alice Harmon

Beginner2022-04-27Added 12 answers

You can always derive Vieta's on the spot
(xp4)(x(p+1))=0
x2(p4+p+1)x+(p+1)(p4)=0
x2x3+k3=0
By comparing coefficient, we have
p4+p+1=13 
5p4=23 
p=815
Now that we have the values of p, can you solve for k?
3(p4)2p4+k=0
3(p+1)2(p+1)+k=0
Subtracting them:
3(p4+p+1)(p4p1)(p4p1)=0
(p4+p+1-13)(p4-p-1)=0
(5p4+23)(-3p4-1)=0
If p=43, then we have p+1=p4=13  and the polynomial should be
3(x+13)2=3x2+2x+13
which is not of the form that is stated.

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