Question

plz solve this question ...

Answer 1

For integer roots $\sf b^2 −4ac$should be a perfect square.

$\sf\to x^{2}−(a+1)x+(a−1)=0$

$\sf\to b^{2}−4ac=(a+1)^2-4(a-1)$

$\sf\to a^{2}−2a+5$

Or

$\sf\to p^2= a^{2}−2a+5$

∴a=1,

Answer 2

required answer :-

for integers roots b²-4ac should be a perfect square.

➡ x²–(a+1)x + (a–1) = 0

➡ b²– 4ac = (a+1)²– 4 (a–1)

➡ a²– 2a + 5

or

➡ p² = a²–2a + 5

= a= 1

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