OG详解-OG9 数学2 Q27

The correct answer is 50. An equation of the form ax²+bx=c, where a, b, and c are constants, has no real solutions if and only if its discriminant, b²-4ac is negative. Applying the distributive property to the left-hand side of the equation x(kx-56)=16 yields kx²-56x=-16. Adding 16 to each side of this equation yields kx²-56x+16=0. Substituting k for a, -56 for b, and 16 for c in b²-4ac yields a discriminant of (-56)² - 4(k)(16) , or 3,136-64k. If the given equation has no real solution, it follows that the value of 3,136-64k must be negative. Therefore, 3 136-64k<0. Adding 64k to both sides of this inequality yields 3,136<64k. Dividing both sides of this inequality by 64 yields 49<k, or k>49. Since it’s given that k is an integer, the least possible value of k is 50.