The number of ordered pairs (a, b) of integer | vedclass

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(B) Given the quadratic equation $x^2+ax+b=0$. Since $a-b$ is a root,it must satisfy the equation:$(a-b)^2 + a(a-b) + b = 0$$a^2 - 2ab + b^2 + a^2 - a

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$4$

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(B) Given the quadratic equation $x^2+ax+b=0$. Since $a-b$ is a root,it must satisfy the equation:$(a-b)^2 + a(a-b) + b = 0$$a^2 - 2ab + b^2 + a^2 - ab + b = 0$$b^2 - 3ab + b + 2a^2 = 0$$b^2 + b(1-3a) + 2a^2 = 0$Solving for $b$ using the quadratic formula:$b = \frac{-(1-3a) \pm \sqrt{(3a-1)^2 - 8a^2}}{2} = \frac{(3a-1) \pm \sqrt{9a^2 - 6a + 1 - 8a^2}}{2} = \frac{(3a-1) \pm \sqrt{a^2 - 6a + 1}}{2}$For $b$ to be an integer,the discriminant $D = a^2 - 6a + 1$ must be a perfect square,say $k^2$.$a^2 - 6a + 1 = k^2$$(a-3)^2 - 8 = k^2$$(a-3)^2 - k^2 = 8$$(a-3-k)(a-3+k) = 8$Since the product is $8$ (even),both factors must be even. Let $X = a-3-k$ and $Y = a-3+k$. Then $XY = 8$ and $X+Y = 2(a-3)$.Possible pairs $(X, Y)$ such that $XY=8$ and $X, Y$ are even:$1$) $X=2, Y=4 \implies 2(a-3)=6 \implies a-3=3 \implies a=6$. Then $k=1$,$b = \frac{(3(6)-1) \pm 1}{2} = \frac{17 \pm 1}{2} \implies b=9, 8$.$2$) $X=-4, Y=-2 \implies 2(a-3)=-6 \implies a-3=-3 \implies a=0$. Then $k=1$,$b = \frac{(3(0)-1) \pm 1}{2} = \frac{-1 \pm 1}{2} \implies b=0, -1$.Thus,the pairs $(a, b)$ are $(6, 9), (6, 8), (0, 0), (0, -1)$.There are $4$ such ordered pairs.

The number of ordered pairs $(a, b)$ of integers such that $a-b$ is a root of $x^2+ax+b=0$ is

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