Product of roots

We have the equation: $$12x^{2}-20x+3\lambda=0, \lambda\in Z$$

Sum of the roots: $$\alpha+\beta\ =-\dfrac{b}{a}=-\ \dfrac{\left(-20\right)}{12}=\dfrac{5}{3}$$

Product of the roots: $$\alpha\ \beta\ =\dfrac{c}{a}=\dfrac{3λ}{12}=\dfrac{λ}{4}$$

Now, we know that: $$\left(\beta\ -\alpha\ \right)^2=\left(\beta\ +\alpha\ \right)^2-4\alpha\ \beta\ $$

$$\left(\beta\ -\alpha\ \right)^2=\left(\dfrac{5}{3}\right)^2-4\left(\dfrac{λ}{4}\right)$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25}{9}-λ$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25-9λ}{9}$$

Taking the square root of both sides:

$$\left|\beta\ -\alpha\right|=\dfrac{\sqrt{\left(25-9λ\right)}}{3}$$

Now, we are given that: 

$$\dfrac{1}{2}\le\ \left|\beta\ -\alpha\right|\le\ \dfrac{3}{2}$$

Substituting:

$$\dfrac{1}{2}\le\ \dfrac{\sqrt{\left(25-9λ\right)}}{3}\le\dfrac{3}{2}$$

Multiplying by 3:

$$\dfrac{3}{2}\le\ \sqrt{\left(25-9λ\right)}\le\dfrac{9}{2}$$

Or, $$\dfrac{9}{4}\le\ 25-9λ\le\dfrac{81}{4}$$

Subtracting 25 from all three:

$$-\dfrac{91}{4}\le\ -9λ\le-\dfrac{19}{4}$$

Dividing all three by -9:

$$\dfrac{91}{36}\ge\ \ λ\ge\dfrac{19}{36}$$

Or, we can say, $$0.527 \leq \lambda \leq 2.527$$

The only integers that fall within the range [0.527, 2.527] are: 1 and 2. 

Hence, the sum of all possible values of $$\lambda$$ is 3.

$$(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$$

$$(x-4)(x-5)=x^2-9x+20$$

Let $$x^2-9x+20=t$$

$$\left(t-9\right)^2-t=3$$

$$t^2-19t+78=0$$

$$\left(t-13\right)\left(t-6\right)=0$$

So, $$t=13\ ,\ 6$$

Now, two cases will be formed: 

Case 1: $$t=13$$

$$x^2-9x+20=13$$

$$x^2-9x+7=0$$

The discriminant is $$\sqrt{53}$$. So, all the roots will be irrational. 

Case 2: $$t=6$$

$$x^2-9x+20=6$$

$$x^2-9x+14=0$$

$$\left(x-7\right)\left(x-2\right)=0$$

$$x=2\ ,\ 7$$

These roots are rational. 

We have to find the product of rational roots. 

Product = $$\left(2\times7\right)=14$$

$$\therefore\ $$ The required answer is A.

We start with the given quadratic expression that involves the modulus function:

$$9x^2-18|x|+5=0.$$

The absolute value $$|x|$$ creates two separate algebraic cases, because by definition

$$ |x|= \begin{cases} x & \text{when } x\ge 0,\\[4pt] -x & \text{when } x<0. \end{cases} $$

We therefore analyse each case separately and afterwards retain only those roots that satisfy the sign condition assumed while forming that case.

Case 1 : We assume $$x\ge 0,$$ so $$|x|=x.$$ Substituting this in the original equation gives

$$9x^2-18x+5=0.$$

To solve any quadratic of the form $$ax^2+bx+c=0,$$ we use the quadratic-formula

$$x=\dfrac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}.$$

Here, $$a=9,\; b=-18,\; c=5.$$ The discriminant is computed first:

$$\Delta=b^2-4ac=(-18)^2-4(9)(5)=324-180=144.$$

Its square-root is $$\sqrt{144}=12.$$ Substituting in the formula:

$$ x=\dfrac{-(-18)\pm 12}{2\cdot 9} =\dfrac{18\pm 12}{18}. $$

This gives two roots

$$x_1=\dfrac{18+12}{18}=\dfrac{30}{18}=\dfrac{5}{3},\qquad x_2=\dfrac{18-12}{18}=\dfrac{6}{18}=\dfrac{1}{3}.$$

Both numbers are non-negative, so both belong to the present case and are valid.

Case 2 : We assume $$x<0,$$ so $$|x|=-x.$$ Substituting this into the original equation yields

$$9x^2-18(-x)+5=9x^2+18x+5=0.$$

Again we identify $$a=9,\; b=18,\; c=5$$ and apply the quadratic formula. First the discriminant:

$$\Delta=b^2-4ac=18^2-4(9)(5)=324-180=144,$$

and again $$\sqrt{\Delta}=12.$$ Thus

$$ x=\dfrac{-18\pm 12}{2\cdot 9} =\dfrac{-18\pm 12}{18}. $$

This furnishes

$$x_3=\dfrac{-18+12}{18}=\dfrac{-6}{18}=-\dfrac{1}{3},\qquad x_4=\dfrac{-18-12}{18}=\dfrac{-30}{18}=-\dfrac{5}{3}.$$

Both are negative, matching the assumption $$x<0,$$ hence both are admissible.

Collecting the four valid roots, we have

$$x_1=\dfrac{5}{3},\; x_2=\dfrac{1}{3},\; x_3=-\dfrac{1}{3},\; x_4=-\dfrac{5}{3}.$$

The product of all roots is

$$ x_1x_2x_3x_4 =\left(\dfrac{5}{3}\right)\!\left(\dfrac{1}{3}\right)\! \left(-\dfrac{1}{3}\right)\!\left(-\dfrac{5}{3}\right). $$

First multiply the two negative factors:

$$\left(-\dfrac{1}{3}\right)\!\left(-\dfrac{5}{3}\right)=\dfrac{5}{9}.$$

Next multiply the two positive factors:

$$\left(\dfrac{5}{3}\right)\!\left(\dfrac{1}{3}\right)=\dfrac{5}{9}.$$

Finally multiply these intermediate results:

$$\dfrac{5}{9}\times\dfrac{5}{9}=\dfrac{25}{81}.$$

Hence, the correct answer is Option B.