SOLUTION: find the values of p for which (7p+1)x^2+(5p-1)x+p=1 has equal roots... thank you

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Question 997509: find the values of p for which (7p+1)x^2+(5p-1)x+p=1 has equal roots...
thank you
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
find the values of p for which %287p%2B1%29x%5E2%2B%285p-1%29x%2Bp=1 has equal roots
For the quadratic equation to have equal roots, the argument to the square root must equal zero and the solution is -b+%2F+2a.
%287p%2B1%29x%5E2%2B%285p-1%29x%2Bp=1
%287p%2B1%29x%5E2%2B%285p-1%29x%2Bp-1=0
b%5E2+-+4ac+=+0+
a=%287p%2B1%29
b=%285p-1%29
c=p-1
b%5E2+-4ac+=+0
%285p-1%29%5E2+-4%287p%2B1%29%28p-1%29+=+0
25p%5E2+-10p%2B1-%2828p%5E2+%2B4%29%28p-1%29+=+0
25p%5E2+-10p%2B1-%2828p%5E3-28p%5E2%2B4p-4%29+=+0
25p%5E2+-10p%2B1-28p%5E3%2B28p%5E2-4p%2B4+=+0
-3p%5E2%2B14p%2B5+=+0...........using quadratic formula, you have
p=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
p=+%28-14+%2B-+sqrt%28+14%5E2-4%2A%28-3%29%2A5+%29%29%2F%282%2A%28-3%29%29+
p=+%28-14+%2B-+sqrt%28+196%2B60+%29%29%2F-6+
p=+%28-14+%2B-+sqrt%28+256+%29%29%2F-6+
p=+%28-14+%2B-+16%29%2F-6+
p=+%28-cross%2814%297+%2B-+cross%2816%298%29%2Fcross%28-6%29-33+
p=+%28-7+%2B-+8%29%2F-3+
solutions:
p=+%28-7+%2B+8%29%2F-3+ =>p=+-1%2F3+
or
p=+%28-7+-+8%29%2F-3+=>p=+-15%2F-3+=>p=+5

then your equation will be
if+p+=-1%2F3
%287%28-1%2F3%29%2B1%29x%5E2%2B%285%28-1%2F3%29-1%29x%2B%28-1%2F3%29-1=0
%28-7%2F3%2B3%2F3%29x%5E2%2B%28-5%2F3-3%2F3%29x-1%2F3-3%2F3=0
-%284%2F3%29x%5E2-%288%2F3%29x-4%2F3=0 => this works, gives us one root
check:
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+-%284%2F3%29x%5E2-%288%2F3%29x-4%2F3%29+

or
if p+=5
%287%285%29%2B1%29x%5E2%2B%285%285%29-1%29x%2B5-1=0
36x%5E2%2B24x%2B4=0=> this works, gives us one root


check:
+graph%28+600%2C+600%2C+-5%2C+5%2C+-10%2C+10%2C+36x%5E2%2B24x%2B4%29+