Quadratic Equation and Inequalities

Given : $p x^2+q x-r=0$ Let $p=\dfrac{a}{r_1}, q=a, r=a r_1$ $$\begin{aligned} & \text{ and } \dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{3}{4} \\\\ & \Rightarrow \dfrac{\alpha+\beta}{\alpha \beta}=\dfrac{3}{4} \\\\ & \Rightarrow \dfrac{-\dfrac{q}{p}}{-\dfrac{r}{p}}=\dfrac{3}{4} \\\\ & \Rightarrow \dfrac{q}{r}=\dfrac{3}{4} \\\\ & \Rightarrow \dfrac{1}{r_1}=\dfrac{3}{4} \\\\ & \Rightarrow r_1=\dfrac{4}{3}\end{aligned}$$ $$\begin{aligned}(\alpha-\beta)^2 & =(\alpha+\beta)^2-4 \alpha \beta \\\\ & =\left(\dfrac{-q}{p}\right)^2-4\left(\dfrac{-r}{p}\right) \\\\ & =\dfrac{q^2}{p^2}+\dfrac{4 r}{p} \\\\ & =r_1^2+4 r_1^2=5 r_1^2 \\\\ & =5\left(\dfrac{4}{3}\right)^2=\dfrac{80}{9}\end{aligned}$$

Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because : $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$ Using the Reciprocal Property : This means $(\sqrt{3} - \sqrt{2})^x = \dfrac{1}{(\sqrt{3} + \sqrt{2})^x}$ Let $a = (\sqrt{3} + \sqrt{2})^x$.

The equation given in the problem becomes : $a + \dfrac{1}{a} = 10$ Simplifying Multiplying both sides by *a*, we get a quadratic equation : $a^2 + 1 = 10a$ $a^2 - 10a + 1 = 0$ Solving the Quadratic Using the quadratic formula, we find : $a = \dfrac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$ Possible values of x Since $a = (\sqrt{3} + \sqrt{2})^x$, we have two cases : $(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$ = $(\sqrt{3} + \sqrt{2})^2$.

There is one real solution for x in this case which is x = 2.

$(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$ = $(\sqrt{3} - \sqrt{2})^2= \dfrac{1}{(\sqrt{3} + \sqrt{2})^2}$ = $(\sqrt{3} + \sqrt{2})^{-2}$.

There is one real solution for x in this case which is x = - 2.

Conclusion There are two real solutions for *x*.

Therefore, the number of elements in the set S is 2.

This corresponds with option (C).

Consider $\dfrac{1}{\sqrt{\mathrm{x}}}=\alpha \quad \mathrm{x}>0$

So, no. of solutions $=4$

$\alpha$ and $\beta$ are the roots of the equation Using Newton's theorem

Put

Put the value of

&

in above equation.

Equation whose roots are

and

is

Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.

The given quadratic equation is:

By Vieta's formulas, the sum of the roots is:

So:

And the product of the roots is:

Therefore, $b = -8a = -8 \left( \dfrac{1}{12} \right) = -\dfrac{2}{3}$. Given the roots of the new quadratic equation are:

and

We know $a = \dfrac{1}{12}$ and $b = -\dfrac{2}{3}$, so:

and:

Thus, the roots of the new quadratic equation are:

and:

The new quadratic equation with roots $-2$ and $-6$ can be formulated as:

The sum of the roots is:

The product of the roots is:

Thus, the quadratic equation becomes:

Hence, the correct option is: Option A

First, let's start by substituting

in the given equation. By substituting, the equation

will be transformed into

Now, we have a quadratic equation in

. To find the roots of this quadratic equation, we can use the quadratic formula:

In this case,

,

, and

. Substituting these values into the formula, we get:

Solving for the two possible values of

, we have:

Now, recall that we substituted

. So, we need to solve for

when

and

:

Therefore, the solutions for

are

and

. The sum of these solutions is:

Using the logarithmic property that

, we get:

Now, we note that:

Thus,

Therefore, the sum of all the solutions of the equation is: Option A

.

$$\begin{aligned} & (a-5)^2-8(15-3 a)