If A and B are the zeros of a quadratic polynomial p(x)=px sq- qx
+r then, a+b-ab is
Answers
Answer:
Step-by-step explanation:
et 2 zeroes be a and b of polynomial x² + px + q = 0
sum of roots = a + b = -p/1 = -p
products of roots = ab = q/1 = q
(a + b)² = a² + b² + 2ab ⇒ a² + b² = p² - 2q
(a-b)² = a² + b² -2ab = p² - 2q -2q = p² - 4q
now ques asks for new quadratic eq whose roots are (a+b) ² and (a-b)²
so sum of new roots are = (a +b)² + (a-b)² = p² + p² -4q = 2p² - 4q
and product of roots = (a+b)²(a-b)² = (p²)² (p²-4q)² = p⁴ (p⁴ +16q² + 8p²q)
hence new quadratic eq gonna be =
x² - x(sum of roots) + (products of roots)
x² - x(2p² - 4q) + p⁴(p⁴ + 16q² +8p²q) = 0
further u can simplify.
hope my ans is correct ...........doubts , then enquire.
Click to let others know, how helpful is it
3.0
100 votes
THANKS 149
Report
Ahmed8486Ambitious
We have polynomial f ( x ) = x2 + p x + q
And
Roots are α and β
And
we know from relationship between zeros and coefficient .
Sum of zeros = −Coefficient of xCoefficient of x2
So,
α + β = - p ------ ( 1 )
Taking whole square on both hand side , we get
( α + β )2 = p 2 ------ ( 2 )
⇒α2 + β2 + 2 α β = p2⇒α2 + β2 + 2 α β − 2αβ + 2αβ= p2⇒α2 + β2 − 2 α β +4 αβ= p2⇒(α − β) 2 +4 αβ= p2 −−−− ( 3 )
And
Products of zeros = Constant termCoefficient of x2
So,
α β = q , Substitute that value in equation 3 , we get
⇒(α− β)2 + 4 (q ) = p2⇒(α− β)2 +4 q= p2⇒(α− β)2 = p2 − 4 q −−−− ( 4 )
Now we add equation 2 and 4 and get
(α + β)2 + (α − β)2 = p2 + p2 − 4 q= 2 p2 − 4 q
And we multiply equation 2 and 4 and get
(α + β)2 × (α − β)2 = p2( p2 − 4 q)= p4 − 4 p2q
And we know formula for polynomial when sum of zeros and product of zeros we know :
Polynomial = k [ x2 - ( Sum of zeros ) x + ( Product of zeros ) ] , Here k is any non zero real number.
Substitute values , we get
Quadratic polynomial = k [ x2 - ( 2 p2 - 4 q) x + ( 2 p4 - 4 p2q) ]
= x2 - ( 2 p2 - 4 q) x + ( 2 p4 - 4 p2q) [ taking k = 1 ] ( Ans )
Hope this information will clear your doubts about topic.