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Let k be the largest integer such that the equation (x−1)2+2kx+11=0(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+9yk/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 + 2kx + 11 = 0

(x−1)2=x2−2x+1(x-1)^2 = x^2 - 2x + 1

So our equation becomes:

x2−2x+1+2kx+11=0x^2 - 2x + 1 + 2kx + 11 = 0

Combining like terms:

x2+x(2k−2)+12=0x^2 + x(2k - 2) + 12 = 0

We need the quadratic in standard form ax2+bx+c=0ax^2 + bx + c = 0 to apply the discriminant condition.


For a quadratic ax2+bx+c=0ax^2 + bx + c = 0 to have no real roots, its discriminant must be negative:

Δ=b2−4ac<0\Delta = b^2 - 4ac < 0

In our case: a=1a = 1, b=(2k−2)b = (2k-2), c=12c = 12

So: (2k−2)2−4(1)(12)<0(2k-2)^2 - 4(1)(12) < 0

(2k−2)2=4k2−8k+4(2k-2)^2 = 4k^2 - 8k + 4

Therefore: 4k2−8k+4−48<04k^2 - 8k + 4 - 48 < 0

4k2−8k−44<04k^2 - 8k - 44 < 0

Dividing by 4: k2−2k−11<0k^2 - 2k - 11 < 0


We need to solve k2−2k−11<0k^2 - 2k - 11 < 0

Complete the square:

k2−2k−11=(k2−2k+1)−1−11=(k−1)2−12k^2 - 2k - 11 = (k^2 - 2k + 1) - 1 - 11 = (k-1)^2 - 12

So: (k−1)2−12<0(k-1)^2 - 12 < 0

(k−1)2<12(k-1)^2 < 12

Taking square roots: ∣k−1∣<12=23≈3.46|k-1| < \sqrt{12} = 2\sqrt{3} \approx 3.46

This gives us: −23<k−1<23-2\sqrt{3} < k-1 < 2\sqrt{3}

1−23<k<1+231-2\sqrt{3} < k < 1+2\sqrt{3}

−2.46<k<4.46-2.46 < k < 4.46

Since we want the largest integer k, we get k=4k = 4.

Quick check: Does k=4k = 4 work? (4−1)2=9<12(4-1)^2 = 9 < 12

Does k=5k = 5 work? (5−1)2=16>12(5-1)^2 = 16 > 12


Now we need to minimize k4y+9y\frac{k}{4y} + 9y where k=4k = 4 and y>0y > 0.

This becomes: 44y+9y=1y+9y\frac{4}{4y} + 9y = \frac{1}{y} + 9y

This is a perfect setup for the AM-GM inequality.

For positive numbers aa and bb:

a+b2≥ab\frac{a + b}{2} \geq \sqrt{ab}

Therefore: a+b≥2aba + b \geq 2\sqrt{ab}

Let us apply this with a=1ya = \frac{1}{y} and b=9yb = 9y:

1y+9y≥21y⋅9y=29=2×3=6\frac{1}{y} + 9y \geq 2\sqrt{\frac{1}{y} \cdot 9y} = 2\sqrt{9} = 2 \times 3 = 6

Equality occurs when a=ba = b, so:

1y=9y\frac{1}{y} = 9y

1=9y21 = 9y^2

y2=19y^2 = \frac{1}{9}

y=13y = \frac{1}{3} (since y>0y > 0)


The least possible value of k4y+9y\frac{k}{4y} + 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 ax+bx\frac{a}{x} + bx.

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