Quadratic Equations

‘a’ and ‘b’ are roots of to find
Considering one of the root ‘ ’ for the equation;

⇒
∴ [Here, consider as ]


Hence, The correct answer is the option (D) 51.

Given:

We can rewrite this as:

Required Expression:
Simplifying further:

Additionally, we know:

To solve this problem, we need to evaluate the expression:

Given that are roots of the quadratic equation:

Using Vieta's formulas, we have:

• Sum of the roots:
• Product of the roots:

The sequence satisfies the recurrence relation:

Substitute the known values:

This recurrence allows us to express any term in the sequence in terms of the previous two terms:

We aim to evaluate:

Substitute the recurrence relations for , , and :

Simplifying these, obtain the combinations back to and .

Now substitute into the expression:

Group terms to factor out and , then simplify to the desired result:

After computations, it reduces to:

Thus, the expression is equal to the option .

To solve the given quadratic expression with a root , where the parameters satisfy , we need to analyze the statements given in the question.

1. Statement Analysis:
   • Statement I: If , then cannot be the geometric mean of and .
   • Statement II: If , then may be the geometric mean of and .
2. Geometric Mean Condition:
   • For to be the geometric mean of and , we have:
3. Quadratic Roots:
   • The quadratic equation is given as .
   • If the quadratic is to have as a root, we look at the signs dependent on parameter values and conditions.
   • If holds true, verify how it affects the root .
4. Checking Statement I:
   • For , assume .
   • If is negative, the polarity of combinations in quadratic terms changes which does not satisfy the geometric mean condition across .
   • Thus, Statement I is true.
5. Checking Statement II:
   • For , again check the root values under .
   • If is positive and not equal to 1, statement for possibility under the geometric mean holds under subsets of parameter configurations.
   • Thus, Statement II is true.

Therefore, the conclusion of this analysis is that both Statement I and II are true.

The Correct answer is option (A): .

To solve the given integral equation, we need to understand and simplify the expression and find the product . The integral is:

Firstly, examine the expression under the integral. Simplifying or transforming parts of the integral might make it easier to evaluate or compare with the right-hand side:

1. Identify potential substitutions or properties of trigonometric functions that could simplify the expression.
2. Use identities such as:
3. Notice that the expression has a common structure that matches with hyperbolic substitutions or trigonometric identities involving sum-to-product or half-angle formulas.

Let's simplify using appropriate substitutions:

• Consider appropriate trigonometric substitutions or transformations to simplify the integrand.
• Check if this can transform to standard integral forms whose solutions are known.

Upon simplifying, you can express the solution components separately, leading to the following steps:

1. Match components of the standard integral form of derivatives.
2. Compute constants and using boundary conditions or known integral solutions: Rewrite expressions and solve for , by setting coefficients equal.
3. Derive expressions from comparing decomposed integrals.

Finally, calculate using the derived values:

Therefore, the correct answer for is .

The problem provides a system of three linear equations in variables with two parameters, and . We are told that the system has infinitely many solutions, and we need to find the value of .

Concept Used:

For a system of linear equations of the form , where is the coefficient matrix and is the constant vector, the condition for having infinitely many solutions is that the determinant of the coefficient matrix is zero, and the determinants of the matrices formed by replacing a column of with are also zero. In terms of Cramer's rule notation, this means:

and

where are the determinants of the matrices obtained by replacing the x, y, and z columns of with the constant vector, respectively.

Step-by-Step Solution:

Step 1: Write down the determinants and .

The given system of equations is:

The determinant of the coefficient matrix is:

The determinant is obtained by replacing the second column (coefficients of y) with the constant terms:

Since does not contain the parameter , it is advantageous to evaluate it first to find .

Step 2: Set and solve for .

For the system to have infinitely many solutions, we must have .

Expanding along the first row:

Step 3: Set and substitute the value of to solve for .

For infinitely many solutions, we must also have .

Expanding along the first row:

Now, substitute the value into this equation:

Step 4: Calculate the value of .

We have found and .

The value of is 38.

The problem provides a quadratic equation with roots and . A sequence is defined as . We are asked to find the value of a given expression involving terms of this sequence. It is a common pattern in such problems that the expression is designed to simplify using a recurrence relation derived from the quadratic equation. The given expression appears to contain a typo, as does not align with the coefficient in the quadratic equation. The intended expression for a clean solution is almost certainly . We will proceed by solving this corrected version.

Concept Used:

If and are the roots of the quadratic equation , they must satisfy the equation. This property can be used to establish a linear recurrence relation for the sequence . Specifically, multiplying the equation by and by and then adding them yields the relation .

Step-by-Step Solution:

Step 1: Establish the equations satisfied by the roots and .

Since and are the roots of , they satisfy the equation:

Step 2: Derive the recurrence relation for the sequence .

Multiply the first equation by and the second equation by (for ):

Adding these two equations together:

Using the definition , we get the recurrence relation:

Step 3: Apply the recurrence relation for .

Substitute into the recurrence relation:

Rearranging this equation gives a direct relationship between the terms in the numerator of the expression we want to evaluate:

Step 4: Substitute the result from the recurrence into the expression.

The expression to be evaluated (with the likely correction) is:

Substitute the expression for the numerator that we found in Step 3:

Step 5: Calculate the final value.

Assuming , we can cancel the term from the numerator and the denominator:

The value of the expression is 4.

To find the quadratic equation whose roots are and , we start by analyzing the roots and given by the equation .

From the given quadratic equation, we know:

• The sum of roots .
• The product of roots .

Using these, we can find powers of and :

Finding

First, compute :

Then, calculate using:

Finding

We can find using the identity:

Now, remember the requirement for the quadratic roots:

The quadratic equation formed by these roots is:

• Sum of roots:
• Product of roots:

The quadratic equation is thus:

Therefore, the correct option is .

Given Roots and Sum/Product Relations:
Since and are roots of the equation , we know:


From the product, we get .

Finding :
Substitute into the sum of roots equation:

Constructing the New Quadratic Equation:
The roots of the new quadratic equation are and .

Substitute and :

Thus, the roots of the new equation are and .

Forming the Equation with Roots and :
A quadratic equation with roots and is:

Given:

Step 1:

Consider the triangle with sides :

Since , Substituting :

Hence,

Case 1:

This is always positive.

Case 2:

Now:

Solving:

Hence,

Now:

Step 2:

∴ Final Answer:

Given that and are the roots of the quadratic equation , where , and it is known that and form consecutive terms of a non-constant Geometric Progression (G.P.). Additionally, it is provided that . We are to find the value of .

Let's start with the properties of the roots of a quadratic equation. For the equation , the roots and satisfy:

• (Sum of roots)
• (Product of roots)

From the given equation, we have .

Substituting the expressions for and , we get:

Simplifying gives:

Since are in G.P., let's denote the common ratio by . Therefore, we have:

From , we find:

Now, since and , substituting in verifies that:

Solving further , thus, .

For the discriminant part to find given by:

Substituting known values:

Using and :

Thus, the correct value is .

Given inequalities: For the discriminant : Considering both conditions together: Now,

We are given the equation:

Case 1:
The absolute value simplifies to: Simplified equation: Solution:

Case 2:
The absolute value becomes: Simplified equation: Solution:

Sum of Squares Calculation:
Expansion: Simplification:

Final Answer:
The sum of squares of the solutions is .