Question

x2 +2px-q2=0 the seeds of the equation are α and β and p, q are rational numbers but if p2 + q2 is not, then find an equation whose seed is α + β + √α +β​

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 = (k⁴+ q² + 2k²q )/k² - 4q

Discriminant = (k⁴ + q²+ 2k²q - 4k²q ) /k²

We know k² > 0

Now lets consider

k⁴+ q²+ 2k²q - 4k²q = k⁴+ 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.

$@shwetasingh1421977$