Suppose k is any integer such that the equation 2x²+kx+5=0 has no real roots and the equation x²+(k-5)x+1=0 has two distinct real roots for x. Then, the number of possible values of k is:
Question:
Suppose k is any integer such that the equation 2x²+kx+5=0 has no real roots and the equation x²+(k-5)x+1=0 has two distinct real roots for x. Then, the number of possible values of k is:
Suppose k is any integer such that the equation 2x²+kx+5=0 has no real roots and the equation x²+(k-5)x+1=0 has two distinct real roots for x. Then, the number of possible values of k is:
Options
Answer: 9
Explanation:
For 2x²+kx+5=0 to have no real roots: k²-40<0, so -√40<k<√40. For x²+(k-5)x+1=0 to have two distinct real roots: (k-5)²-4>0, so k<3 or k>7. Combining these conditions with k being an integer gives 9 possible values.
Explanation:
For 2x²+kx+5=0 to have no real roots: k²-40<0, so -√40<k<√40. For x²+(k-5)x+1=0 to have two distinct real roots: (k-5)²-4>0, so k<3 or k>7. Combining these conditions with k being an integer gives 9 possible values.
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