The sum of all integral values of k (k ≠ 0) for which the equation 2/(x - 1) - 1/(x - 2) = 2/k in x has no real roots, is ____________.

Solution

Maths sir · 5 steps

Clear the fractions, get a quadratic in x, and ask when its discriminant goes negative.

1. Combine the left side. 2/(x-1)-1/(x-2) = (2(x-2)-(x-1))/(x-1)(x-2) = (x-3)/(x-1)(x-2), with x ≠ 1, 2.
2. Cross-multiply. Setting this equal to 2/k gives k(x-3) = 2(x²-3x+2), which rearranges to 2x²-(6+k)x+(4+3k)=0.

   D = (6+k)² - 8(4+3k) = k² - 12k + 4 < 0

3. Solve the inequality. The roots of k²-12k+4=0 are k = 6 ± 4√2, about 0.34 and 11.66. So D<0 for 6-4√2 < k < 6+4√2, and the integers there are k = 1, 2, …, 11.
4. Check excluded points. Putting x=1 into the quadratic gives 2k=0, and putting x=2 gives k=0. Since k ≠ 0, neither excluded value is ever a root, so no other k loses its roots this way.
5. So. The sum is 1+2+…+11 = (11 × 12)/2 = 66.

The idea to keep. For a rational equation, clear denominators to get a quadratic, use D<0, then check that the excluded values of x cannot sneak in as roots and change the answer.

Answer from the NTA final answer key. Solution by PrepWiser.