If 3 + 2\(\sqrt{2}\) and 3 − 2\(\sqrt{2}\) are the roots of a quadratic equation, then the quadratic equation is:
x² − 6x + 1 = 0
x² − 6x − 1 = 0
x² + 6x + 1 = 0
x² + 6x − 1 = 0
We are given two roots of a quadratic equation. We can use the relationship between the roots and coefficients of a quadratic equation to find the equation. A quadratic equation with roots \(\alpha\) and \(\beta\) is given by \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\), where \((\alpha + \beta)\) is the sum of the roots and \(\alpha\beta\) is the product of the roots.
Identify the roots:
Let the given roots be \(\alpha = 3 + 2\sqrt{2}\) and \(\beta = 3 - 2\sqrt{2}\).
Calculate the sum of the roots:
\[ \alpha + \beta = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) \]\[ \alpha + \beta = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} \]\[ \alpha + \beta = 3 + 3 \]\[ \alpha + \beta = 6 \]
Calculate the product of the roots:
\[ \alpha\beta = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \]This is in the form \((a+b)(a-b) = a^2 - b^2\).
Here, \(a = 3\) and \(b = 2\sqrt{2}\).
\[ \alpha\beta = (3)^2 - (2\sqrt{2})^2 \]\[ \alpha\beta = 9 - (4 \times 2) \]\[ \alpha\beta = 9 - 8 \]\[ \alpha\beta = 1 \]
Form the quadratic equation:
The general form of a quadratic equation with roots \(\alpha\) and \(\beta\) is:
\[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \]Substitute the sum and product of the roots we calculated:
\[ x^2 - (6)x + (1) = 0 \]\[ x^2 - 6x + 1 = 0 \]
Correct Option: A) The calculated quadratic equation is \(x^2 - 6x + 1 = 0\), which matches option A.
The problem requires us to decode a pattern from two given examples and then apply that pattern to a new word. We need to analyze the transformation of each letter in the given words to identify the rule.
The pattern observed is: +1, -2, +1, -2, +1.
This confirms the pattern: +1, -2, +1, -2, +1.
Correct Option: C) KSNNZ
The question asks for the city that hosted the first Khelo India Fencing Women's League in July 2022. This is a factual question related to sports events in India, requiring knowledge of recent national sports initiatives.
Correct Option: B) New Delhi
The first Khelo India Fencing Women's League was inaugurated in New Delhi in July 2022. This league was a significant step by the Ministry of Youth Affairs and Sports to promote women's fencing in India, providing a platform for young female fencers to showcase their talent.
In a row of people standing facing north, Vikram stands 11th from the left end of the row. His friend Aditya stands 3rd from the right end. There are 14 people standing between Vikram and Aditya. How many people are standing in that row?
To find the total number of people in the row, we need to sum the positions of Vikram and Aditya from their respective ends, and the number of people between them. Since Vikram's position is given from the left and Aditya's from the right, and there are people between them, we can directly add these values.
Let's use the formula for total people in a row when positions from both ends and people between are given:
Total people = (Position of 1st person from left) + (Number of people between) + (Position of 2nd person from right)
In this case:
Total number of people \( = \) (Vikram's position from left) \( + \) (Number of people between them) \( + \) (Aditya's position from right)
\[ \text{Total people} = 11 + 14 + 3 \]
\[ \text{Total people} = 28 \]
A) 28 is the correct answer. By adding Vikram's position from the left, the number of people between them, and Aditya's position from the right, we get the total number of people in the row.
The question asks to identify the true statements about land use. We need to evaluate each statement individually to determine its correctness based on general geographical and environmental knowledge.
D) 4, 3 and 1
Based on our evaluation, statements 1, 3, and 4 are true, while statement 2 is false. Therefore, the correct set of true statements is 1, 3, and 4.
This problem involves calculating the work done by pipes filling a cistern. We need to determine the individual filling rates of Pipe A and Pipe B. Since the total time for filling is given, and Pipe B works for the entire duration, we can calculate the portion of the cistern filled by Pipe B. The remaining portion must have been filled by Pipe A (and Pipe B) when both were open. This allows us to find the time Pipe A was open.
Correct Option: A) 8 min
If '+' means '−', '−' means '×', '×' means '÷', '÷' means '+', what will come in place of the question mark (?) in the following equation?
15 − 2 ÷ 90 × 9 + 10 =?
This problem involves substituting given symbol meanings into a mathematical expression and then evaluating the expression according to the order of operations (BODMAS/PEMDAS).
The original equation is: \(15 - 2 \div 90 \times 9 + 10 = ?\)
Applying the substitutions:
The new equation becomes: \(15 \times 2 + 90 \div 9 - 10 = ?\)
BODMAS stands for Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction. Operations are performed from left to right for Division/Multiplication and then for Addition/Subtraction.
First, perform Multiplication and Division from left to right:
Substitute these values back into the equation:
\(30 + 10 - 10 = ?\)
Next, perform Addition and Subtraction from left to right:
So, \(? = 30\).
D) 30. After substituting the operators and applying the BODMAS rule, the expression evaluates to 30.
The HCF of fractions is calculated as \(\frac{\text{HCF of the numerators}}{\text{LCM of the denominators}}\). Find the HCF of \( \frac{2}{3} \), \( \frac{4}{5} \) and \( \frac{3}{2} \).
\( \frac{5}{30} \)
\( \frac{3}{50} \)
\( \frac{1}{30} \)
\( \frac{1}{40} \)
To find the HCF (Highest Common Factor) of fractions, we use the formula: \[\text{HCF of fractions} = \frac{\text{HCF of the numerators}}{\text{LCM of the denominators}}\]. We will apply this formula to the given fractions \( \frac{2}{3} \), \( \frac{4}{5} \) and \( \frac{3}{2} \).
C) \( \frac{1}{30} \) is the correct HCF of the given fractions, as calculated by finding the HCF of the numerators and the LCM of the denominators.
This problem involves solving a system of two equations with two unknowns, representing the series and parallel combinations of two resistors. We will use the formulas for equivalent resistance in series and parallel circuits to set up the equations and then solve for the individual resistances.
Formulate equations based on given information:
When two resistors \(R_1\) and \(R_2\) are connected in series, their net resistance \(R_{\text{series}}\) is given by:
\[ R_{\text{series}} = R_1 + R_2 \]Given that the net resistance in series is 8 ohm, we have:
\[ R_1 + R_2 = 8 \quad \text{(Equation 1)} \]When the same two resistors are connected in parallel, their net resistance \(R_{\text{parallel}}\) is given by:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \]This can be simplified to:
\[ R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2} \]Given that the net resistance in parallel is 2 ohm, we have:
\[ \frac{R_1 R_2}{R_1 + R_2} = 2 \quad \text{(Equation 2)} \]Substitute Equation 1 into Equation 2:
From Equation 1, we know \(R_1 + R_2 = 8\). Substitute this into Equation 2:
\[ \frac{R_1 R_2}{8} = 2 \]Multiply both sides by 8 to find the product of \(R_1\) and \(R_2\):
\[ R_1 R_2 = 16 \quad \text{(Equation 3)} \]Solve the system of equations:
We now have a system of two equations:
1. \(R_1 + R_2 = 8\)
2. \(R_1 R_2 = 16\)
We can express \(R_2\) from Equation 1 as \(R_2 = 8 - R_1\). Substitute this into Equation 3:
\[ R_1 (8 - R_1) = 16 \]Expand the equation:
\[ 8R_1 - R_1^2 = 16 \]Rearrange into a standard quadratic form \(a x^2 + b x + c = 0\):
\[ R_1^2 - 8R_1 + 16 = 0 \]This is a perfect square trinomial, \((R_1 - 4)^2 = 0\). Alternatively, use the quadratic formula or factorization.
Factoring:
\[ (R_1 - 4)(R_1 - 4) = 0 \]Solving for \(R_1\):
\[ R_1 - 4 = 0 \implies R_1 = 4 \]Find \(R_2\):
Substitute the value of \(R_1 = 4\) into Equation 1:
\[ 4 + R_2 = 8 \]Solving for \(R_2\):
\[ R_2 = 8 - 4 \implies R_2 = 4 \]Thus, both resistors have a value of 4 ohm.
Correct Option: B) 4 ohm; 4 ohm
As calculated, the individual resistances are \(R_1 = 4\) ohm and \(R_2 = 4\) ohm.