Let k be the largest integer such that the equation (x−1)2+2kx+11=0 has no real roots. If y is a positive real number, then the least possible value of k/4y+9y is
Entered answer:
Solution
✅ Correct Answer: 6
We need to find the largest integer k for which a quadratic has no real roots, then use that k to minimize an expression.
Starting with: (x−1)2+2kx+11=0
(x−1)2=x2−2x+1
So our equation becomes:
x2−2x+1+2kx+11=0
Combining like terms:
x2+x(2k−2)+12=0
We need the quadratic in standard form ax2+bx+c=0 to apply the discriminant condition.
For a quadratic ax2+bx+c=0 to have no real roots, its discriminant must be negative:
Δ=b2−4ac<0
In our case: a=1, b=(2k−2), c=12
So: (2k−2)2−4(1)(12)<0
(2k−2)2=4k2−8k+4
Therefore: 4k2−8k+4−48<0
4k2−8k−44<0
Dividing by 4: k2−2k−11<0
We need to solve k2−2k−11<0
Complete the square:
k2−2k−11=(k2−2k+1)−1−11=(k−1)2−12
So: (k−1)2−12<0
(k−1)2<12
Taking square roots: ∣k−1∣<12=23≈3.46
This gives us: −23<k−1<23
1−23<k<1+23
−2.46<k<4.46
Since we want the largest integer k, we get k=4.
Quick check: Does k=4 work? (4−1)2=9<12
Does k=5 work? (5−1)2=16>12
Now we need to minimize 4yk+9y where k=4 and y>0.
This becomes: 4y4+9y=y1+9y
This is a perfect setup for the AM-GM inequality.
For positive numbers a and b:
2a+b≥ab
Therefore: a+b≥2ab
Let us apply this with a=y1 and b=9y:
y1+9y≥2y1⋅9y=29=2×3=6
Equality occurs when a=b, so:
y1=9y
1=9y2
y2=91
y=31 (since y>0)
The least possible value of 4yk+9y is 6.
Key insight: No real roots corresponds to a negative discriminant, and AM-GM inequality is perfect for minimizing expressions of the form xa+bx.