The integral values of a for which the quadra | vedclass

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(A) The given quadratic equation is $(x - a)(x - 10) + 1 = 0$.Expanding this,we get $x^{2} - (10 + a)x + 10a + 1 = 0$.For the roots to be integers,the

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$8, 12$

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(A) The given quadratic equation is $(x - a)(x - 10) + 1 = 0$.Expanding this,we get $x^{2} - (10 + a)x + 10a + 1 = 0$.For the roots to be integers,the discriminant $D$ must be a perfect square.$D = b^{2} - 4ac = (10 + a)^{2} - 4(10a + 1)$.$D = 100 + 20a + a^{2} - 40a - 4$.$D = a^{2} - 20a + 96$.Completing the square,$D = (a - 10)^{2} - 4$.Let $D = k^{2}$ for some integer $k \ge 0$.Then $k^{2} = (a - 10)^{2} - 4$,which implies $(a - 10)^{2} - k^{2} = 4$.$(a - 10 - k)(a - 10 + k) = 4$.Since the difference between these two factors is $2k$ (an even number),both factors must be even.The possible pairs of factors of $4$ are $(2, 2)$ or $(-2, -2)$.Case $1$: $a - 10 - k = 2$ and $a - 10 + k = 2$.Adding these,$2(a - 10) = 4 \Rightarrow a - 10 = 2 \Rightarrow a = 12$.Case $2$: $a - 10 - k = -2$ and $a - 10 + k = -2$.Adding these,$2(a - 10) = -4 \Rightarrow a - 10 = -2 \Rightarrow a = 8$.Thus,the integral values of $a$ are $8$ and $12$.

The integral values of $a$ for which the quadratic equation $(x - a)(x - 10) + 1 = 0$ has integral roots are

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