Question

Solve asap and correctly.

Answer 1

Logical Method:

Final Answer :Polynomial : x^2-101x+2500.

Understanding:
1) To get the polynomial whose zeroes are squares of given polynomial. Replace x with root(x)
Steps:
1) Replace x with √x.
2) Then solve.

Calculation:
1) (√x) ^2+√x -50=0
=>x-50 = (-√x)
=> (x-50)^2 = (-√x)^2
=> x^2-100x+2500=x
=> x^2-101x+2500=0
Now, the required Polynomial is
x^2-101x+2500
For Calculation see pic

Answer 2

Hey there!

Required to find a polynomial whose zeroes are squares of the zeroes of x² + x - 50 .

We know that, For a quadratic polynomial in the form of ax² + bx + c, Sum of zeroes and product of zeroes are -b/a , c/a respectively

Now, For the polynomial x² + x - 50 .

Let the zeroes be p, q

p + q = Sum of roots = -1/1 = -1

pq = Product of roots. -50/1 = -50 .

Now,
For the required polynomial ,

Zeroes are p² , q²

Sum of roots = p² + q² = ( p + q) ² -2pq = (-1)² - 2 (-50 ) = 1 + 100 = 101 .

Product of roots = p²*q² = (pq)² = (-50)² = 2500 .

Now,
We know that,
Quadratic polynomial = x² - ( Sum of roots) x + Product of roots.

The required polynomial = x² - ( 101) x + 2500 = x² - 101x + 2500

Hope helped!