If p and q are the zeroes of the polynomial 4x2
-3x +2 ,find the value of
1/p2
+ 1/q2
Answers
Answer:
– 7/4
Note:
If A and B are the zeros of a quadratic polynomial ax² + bx + c , then ;
Sum of zeros , (A+B) = -b/a
Product of zeros , (A•B) = c/a
Solution:
Here,
The given quadratic polynomial is :
4x² - 3x + 2
Clearly,
a = 4
b = -3
c = 2
Also,
It it given that , p and q are the zeros of the given quadratic polynomial.
Thus,
Sum of the zeros = -b/a
=> p + q = -(-3)/4 = 3/4
Also,
Product of zeros = c/a
=> pq = 2/4 = 1/2
Thus,
q² + p²
=> 1/p² + 1/q² = --------------
p²q²
p² + q²
=> 1/p² + 1/q² = --------------
p²q²
( p + q )² - 2pq
=> 1/p² + 1/q² = -------------------------
( pq )²
( 3/4 )² - 2(1/2)
=> 1/p² + 1/q² = -------------------------
( 1/2 )²
9/16 – 1
=> 1/p² + 1/q² = ----------------
1/4
( 9 – 16 )/16
=> 1/p² + 1/q² = ---------------------
1/4
– 7/16
=> 1/p² + 1/q² = -------------
1/4
– 7 × 4
=> 1/p² + 1/q² = -------------
16
=> 1/p² + 1/q² = – 7/4
Hence,
The required value of 1/p² + 1/q² = – 7/4
Given : p & q are zeroes of polynomial 4x² - 3x + 2
To find : Value of 1/p² + 1/q²
Solution:
p & q are zeroes of polynomial
4x² - 3x + 2
Sum of roots = p + q = -(-3)/4 = 3/4
Products of roots pq = 2/4 = 1/2
p + q = 3/4
pq = 1/2
1/p² + 1/q²
= ( q² + p²)/p²q²
=( ( P + q)² - 2pq ) / ( pq )²
= ( (3/4)² - 2(1/2) ) / ( 1/2)²
= ( 9/16 - 1 ) / (1/4)
= ( -7/16 )(1/4)
= -7/4
1/p² + 1/q² = - 7/4
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