Derieving Derivative #6
x 2 − ( a − 2 ) x − ( a + 1 ) = 0
If p, q are the roots of the equation above, then find the least value of p 2 + q 2 .
The answer is 5.
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2 solutions
good solution !!!
Or even, the vertex of the parabola a 2 − 2 a + 6 is a = 1 .
Since it is opening upwards, the minimum value occurs at a = 1 , which is 5.
x 2 − ( a − 2 ) x − ( a + 1 ) = 0 now , p + q = a − 2 and p q = − ( a + 1 ) p 2 + q 2 = a 2 − 4 a + 4 + 2 a + 2 Therefore , f ( a ) = a 2 − 2 a + 6 f ′ ( a ) = 2 a − 2 = 0 since f ′ ′ ( a ) > 0 So a=1 and thus f(a)=5 is the least value for p 2 + q 2
Did the same way,nice solution.
Thanks mate
it can be done by am - gm inequality
Dude, in the very 1st step, p+q=(a-2) not 2, check it..!!
We don't need calculus to find the minimum value.
p 2 + q 2 = a 2 − 2 a + 6 = a 2 − 2 a + 1 + 5 = ( a − 1 ) 2 + 5
We note that ( a − 1 ) 2 is always positive and it has a minimum value of 0 when a = 1 , therefore the minimum value of p 2 + q 2 is 5 .