Question

Show that the equation x^2 + 2px- 3 = 0 has real and distinct roots for all values of p

Answer 1

Answer:

Discriminant of $\tt{x^{2} + 2px - 3=0}$:

$\tt{D = b^{2} - 4ac}$

=> $\tt{D = 4p^{2} + 12}$

Since the square of any positive integer can't be negative, the discriminant will always be greater than 0.

hence proved.

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Answer 2

Step-by-step explanation:

Discriminant of x²+ 2px - 3=0

x²+2px−3=0 :

D = b⁴- 4ac

D=b²−4ac

=>D = 4p² + 12

D=4p²+12

Since the square of any positive integer can't be negative, the discriminant will always be greater than 0.

hence proved.