Quadratic Equation with Parameter

We consider the quadratic equation $$2x^2 + (a-5)x + 15 - 3a = 0$$ and require that it have no real roots, which is equivalent to its discriminant satisfying $$D = (a-5)^2 - 4(2)(15-3a) < 0\;.$$

Expanding the discriminant yields $$a^2 - 10a + 25 - 120 + 24a < 0\;,$$ which simplifies to $$a^2 + 14a - 95 < 0\;.$$

To find the critical values of $$a$$, we solve the equation $$a^2 + 14a - 95 = 0$$, giving

$$a = \frac{-14 \pm \sqrt{196 + 380}}{2} = \frac{-14 \pm \sqrt{576}}{2} = \frac{-14 \pm 24}{2}\;,$$

so that $$a = 5$$ or $$a = -19\;.$$

Therefore the quadratic has no real roots precisely when $$a \in (-19, 5)\;.$$

For integer values of $$x$$ in this interval we set

$$X = \{\,x \in \mathbb{Z} : -19 < x < 5\} = \{-18, -17, \dots, 4\}\;.$$

We wish to compute

$$\sum_{x \in X} x^2 = \sum_{x=-18}^{4} x^2 = \sum_{x=1}^{18} x^2 + \sum_{x=1}^{4} x^2\;.$$

Using the formula $$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\;,$$ we find

$$\sum_{x=1}^{18} x^2 = \frac{18 \times 19 \times 37}{6} = \frac{12654}{6} = 2109$$

and

$$\sum_{x=1}^{4} x^2 = \frac{4 \times 5 \times 9}{6} = 30\;.$$

Hence

$$\sum_{x \in X} x^2 = 2109 + 30 = 2139\;.$$

The correct answer is Option 4: 2139.

We start with the given equation

$$ (3x^2+4x+3)^2-(k+1)(3x^2+4x+3)(3x^2+4x+2)+k(3x^2+4x+2)^2=0,\qquad k\gt -1. $$

Notice that the two quadratic expressions which appear repeatedly differ by exactly $$1$$:

$$ 3x^2+4x+3=(3x^2+4x+2)+1. $$

So we introduce the substitution

$$ A=3x^2+4x+2, \qquad \text{hence } 3x^2+4x+3=A+1. $$

Re-writing every occurrence in terms of $$A$$ gives

$$ (A+1)^2-(k+1)(A+1)A+kA^2=0. $$

Now we expand every product:

$$ (A+1)^2=A^2+2A+1, $$

$$ (A+1)A=A^2+A. $$

Substituting these expansions, the whole equation becomes

$$ (A^2+2A+1)-(k+1)(A^2+A)+kA^2=0. $$

We now collect like terms. First the $$A^2$$ terms:

$$ A^2 - (k+1)A^2 + kA^2 \;=\; \bigl[1-(k+1)+k\bigr]A^2 \;=\; (1-k-1+k)A^2 \;=\; 0\cdot A^2 \;=\;0. $$

Thus the entire $$A^2$$ part cancels out completely.

Next the $$A$$ terms:

$$ 2A-(k+1)A \;=\; (2-k-1)A \;=\; (1-k)A. $$

Finally, the constant term is simply $$+1$$.

So the original rather complicated equation has reduced to the linear equation in $$A$$

$$ (1-k)A+1=0. $$

If $$k\neq1$$, we can solve for $$A$$ straight away:

$$ A=\frac{-1}{1-k}=\frac{1}{k-1}. $$

However, if $$k=1$$ the coefficient in front of $$A$$ becomes zero and we are left with the contradictory statement $$1=0$$. Therefore

$$ k=1 \quad\text{gives no root at all and must be excluded}. $$

For every other value of $$k$$ (still keeping $$k\gt -1$$ in mind) the problem now boils down to asking when the quadratic in $$x$$

$$ 3x^2+4x+2=\frac{1}{k-1} $$

has real solutions. Rearranging this gives a standard quadratic equation:

$$ 3x^2+4x+\Bigl(2-\frac{1}{k-1}\Bigr)=0. $$

For such a quadratic $$ax^2+bx+c=0$$ to have real roots, its discriminant must satisfy $$\Delta=b^2-4ac\ge 0.$$ Here

$$ a=3,\qquad b=4,\qquad c=2-\frac{1}{k-1}. $$

Therefore

$$ \Delta =4^2-4\cdot3\Bigl(2-\frac{1}{k-1}\Bigr) =16-12\Bigl(2-\frac{1}{k-1}\Bigr). $$

Expanding the right-hand side, we get

$$ \Delta =16-24+12\cdot\frac{1}{k-1} =-8+\frac{12}{k-1}. $$

Requiring $$\Delta\ge0$$ gives

$$ -8+\frac{12}{k-1}\ge0 \;\;\Longrightarrow\;\; \frac{12}{k-1}\ge8 \;\;\Longrightarrow\;\; \frac{1}{k-1}\ge\frac{2}{3}. $$

Now we solve the inequality $$\dfrac{1}{k-1}\ge\dfrac{2}{3}$$ carefully, taking into account the sign of the denominator.

Case I: $$k-1\gt 0$$ (that is, $$k\gt 1$$).
Multiplying by the positive quantity $$(k-1)$$ preserves the inequality:

$$ 1\;\ge\;\frac{2}{3}(k-1) \;\;\Longrightarrow\;\; k-1\;\le\;\frac{3}{2} \;\;\Longrightarrow\;\; k\;\le\;\frac{5}{2}. $$

Together with $$k\gt 1$$, this yields the interval

$$ 1\lt k\le\frac{5}{2}. $$

Case II: $$k-1\lt 0$$ (that is, $$k\lt 1$$).
Here the denominator is negative, so multiplying reverses the inequality:

$$ 1\;\le\;\frac{2}{3}(k-1). $$

The right side is negative because $$(k-1)\lt 0$$, while the left side is positive, which is impossible. Thus no solutions arise from this case.

Combining everything, remembering also that $$k=1$$ is excluded and the original condition was $$k\gt -1$$, we obtain the final set of all permissible values of $$k$$:

$$ \,1\lt k\le\frac{5}{2}\,. $$

Looking at the options, this interval is exactly Option B.

Hence, the correct answer is Option B.

We begin with the given equation

$$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k},\qquad k\neq 0.$$

To clear the fractions we first take the common denominator on the left side. The common denominator of the two fractions is $$(x-1)(x-2).$$ Hence

$$\frac{2}{x-1}-\frac{1}{x-2}\;=\;\frac{2(x-2)-(x-1)}{(x-1)(x-2)}.$$

Expanding the numerator we have

$$2(x-2)-(x-1)=2x-4-x+1=x-3.$$

Therefore the equation becomes

$$\frac{x-3}{(x-1)(x-2)}=\frac{2}{k}.$$

Now we cross-multiply (which is allowed because none of the denominators is zero when a solution is valid):

$$k(x-3)=2\,(x-1)(x-2).$$

Next we expand the right-hand side. First note that

$$(x-1)(x-2)=x^2-3x+2.$$

Multiplying by the factor $$2$$ we get

$$2(x-1)(x-2)=2x^2-6x+4.$$

So the cross-multiplied equation is

$$k(x-3)=2x^2-6x+4.$$

We now bring every term to one side to form a quadratic equation in $$x$$. Distributing $$k$$ on the left gives

$$kx-3k=2x^2-6x+4.$$

Subtracting the left-hand side from both sides gives

$$0=2x^2-6x+4-kx+3k.$$

Grouping like terms in $$x$$ we write

$$0=2x^2-(k+6)x+(3k+4).$$

Thus the quadratic in standard form is

$$2x^2-(k+6)x+(3k+4)=0.$$

The quadratic formula tells us that a quadratic equation $$ax^2+bx+c=0$$ has real roots precisely when its discriminant $$\Delta=b^2-4ac$$ is non-negative. We want no real roots, so we require $$\Delta\lt 0.$$

Here $$a=2,\; b=-(k+6),\; c=3k+4.$$ Hence

$$\Delta=(-(k+6))^2-4\cdot2\cdot(3k+4).$$

Calculating each part we get

$$(k+6)^2=k^2+12k+36,$$

and

$$4\cdot2\cdot(3k+4)=8(3k+4)=24k+32.$$

Therefore

$$\Delta=k^2+12k+36-(24k+32)=k^2-12k+4.$$

For no real roots we need

$$k^2-12k+4\lt 0.$$

This is a quadratic inequality in $$k$$. Because the coefficient of $$k^2$$ is positive, the expression is negative between its two real roots. We find the roots by setting the quadratic equal to zero:

$$k^2-12k+4=0.$$

Using the quadratic formula with respect to $$k$$ (now $$a=1,\; b=-12,\; c=4$$) we get

$$k=\frac{12\pm\sqrt{(-12)^2-4\cdot1\cdot4}}{2}=\frac{12\pm\sqrt{144-16}}{2}=\frac{12\pm\sqrt{128}}{2}=\frac{12\pm8\sqrt2}{2}=6\pm4\sqrt2.$$

Thus the inequality $$k^2-12k+4\lt 0$$ holds for

$$6-4\sqrt2\lt k\lt 6+4\sqrt2.$$

Since $$4\sqrt2\approx5.656854,$$ the numerical bounds are approximately

$$0.343146\lt k\lt 11.656854.$$

We are asked for integral (integer) values of $$k$$ with the additional condition $$k\neq0$$. The integers that satisfy the inequality are therefore

$$k=1,2,3,4,5,6,7,8,9,10,11.$$

Adding these values we obtain

$$1+2+3+4+5+6+7+8+9+10+11=\frac{11\,(1+11)}{2}=66.$$

So, the answer is $$66$$.