Question

If a and b are the zeroes of the polynomial
f(x) = x^2-px+2 then find the value of a^2+b^2​

Answer 1

Solution:-

Roots of Polynomial F(x)= a and b...given

Polynomial F(x) = x²-px+2

TO FIND :-

=>a²+b²

STEPS TO SOLVE:-

Since,

a²+b²= (a+b)²-2ab

Therby,

=> a+b = p/1

=> a.b = 2/1

So, likewise

=> a²+b²=(a+b)²-2ab

=> (p)²-2(2)

=> p²-4

So,

value of (a²+b²)is this given polynomial is =>P²-4

General points:-

Any equation is of the form ,

F(x)= x²-px+2

Is,

=> x²-(a+b)x+ab=0

Sum of zeroes and product of zeroes define the, Polynomial, and does not matter with any other points.

What are zeroes?

Zeroes are thoese values or equations which turn the given major equation to, 0 or ends. it.

Ans.

Answer 2

Answer :-

a² + b² = p² - 4

Solution :-

f( x ) = x² - px + 2

a and b are the zeroes of the polynomial

Comparing the polynomial with ax² + bx + c we get,

• a = 1
• b = - p
• c = 2

Sum of zeroes = a + b = - b/a = - ( - p ) / 1 = p

⇒ a + b = p

Product of zeroes = c/a = 2/1 = 2

⇒ ab = 2

Using identity ( a + b )² = a² + b² + 2ab

⇒ ( a + b )² = a² + b² + 2ab

⇒ ( p )² = a² + b² + 2( 2 )

⇒ p² = a² + b² + 4

⇒ p² - 4 = a² + b²

⇒ a² + b² = p² - 4

Hence, the value of a² + b² is p² - 4.