Question

(i) Prove that if the roots of x^2 -2px +q =0 are equal then the roots of (1+y)x^2 -2(p+y)x +(q+y) =0 will be real and unequal provided y<0 and p is not equal to 1. (ii) of p=k+q/k then prove that the roots of x^2 + Px + q=0 are rational, p, q, K being rationals.​

Answer 1

Given:

( i ) The roots of x² -2px +q =0 are equal.

( ii ) for x² -2px +q =0 p=k+q/k

To Prove:

( i ) Roots of (1+y)x^2 -2(p+y)x +(q+y) =0 will be real and unequal provided y<0 and p is not equal to 1.

( ii ) Roots of x^2 + Px + q=0 are rational, p, q, K being rationals.​

Solution:

          ( i ) Given roots of  x² -2px + q = 0 are equal .

• Discriminant = 0
• b² -4ac = 0
• 4p² - 4q = 0
• p² = q - ( 1 )
• Let f ( x ) = x² -2px + q = x² -2px + p²
• f(1 ) = 1 -2p + p²  = ( p + 1 )² > 0

We have other equation :  (1+y)x² -2(p+y)x +(q+y) =0  

• Roots if this equation :
• x =( -b ± √b² -4ac)/2a
• x =  ( 2(p+y) ± √ 4(p+y)² - 4(1+y)(q+y) ) /2(1+y)
• x =  ( (p+y) ± √(p+y)² - (1+y)(p² +y))/(1+y)

For real and  unequal roots Discriminant > 0

• (p+y)² - (1+y)(p² +y) > 0
• p² +2py + y² - p² - y - p²y - y² = 2py - y - p²y
• Discriminant = y ( 2p - p² -1 ) = y x - ( p² -2p + 1 )
• Discriminant = y x  - ( p + 1 )²
• Y < 0  and (p + 1 )² > 0 as it is a square .
• Hence Discriminant is positive .
• Therefore roots are real and unequal.

   ( i i )  Given p = k + q/k and f(x) = x² + px + q .

• For rational roots, Discriminant ≥ 0 .
• Therefore p² -4q ≥ 0
• Given p = k + q/k  = (k² + q)/k
• (k²+q)²/k² - 4q = ($k^{4}$ + q² + 2k²q )/k² - 4q
• Discriminant = ($k^{4}$ + q²+ 2k²q - 4k²q ) /k²
• We know k² > 0

Now lets consider

• $k^{4}$ + q²+ 2k²q - 4k²q  = $k^{4}$ + q² - 2k²q = ( k² - q)²
• Since it is a square number , ( k² - q )² ≥ 0
• Therefore , Discriminant ≥ 0
• Therefore roots of x² + px + q = 0 are rational.

Thus proved that if the roots of x² -2px +q =0 are equal then the roots of (1+y)x² -2(p+y)x +(q+y) =0 will be real and unequal provided y<0 and p is not equal to 1.

Also proved that if p=k+q/k that the roots of x² + Px + q=0 are rational, p, q, K being rationals.​