64x²-(16a+4b)x+ab=0 In the Given Equation, a and b are Positive Constants. Find the Value of k

64x²-(16a+4b)x+ab=0 In the given equation, a and b are positive constants. The sum of the solutions to the given equation is k(4a+b), where k is a constant. What is the value of k ?

ANSWER:

$\frac{1}{16}$

Explanation:

Step 1: Understanding the Sum of Solutions Formula

For any quadratic equation of the form $ax^2 + bx + c = 0$, the sum of its solutions is given by the formula:

$$-\frac{b}{a}$$

Step 2: Identifying Coefficients in the Given Equation

The given equation is:

$$64x^2 - (16a + 4b)x + ab = 0$$

Here:

• The coefficient of $x^2$ (which is $a$ in the general formula) is $64$.
• The coefficient of $x$ (which is $b$ in the general formula) is $-(16a + 4b)$.

Step 3: Calculating the Sum of Solutions

Using the formula for the sum of solutions:

$$-\frac{-(16a + 4b)}{64} = \frac{16a + 4b}{64}$$

Simplify this expression:

$$\frac{16a + 4b}{64} = \frac{4a + b}{16}$$

Step 4: Equating the Expression with $k(4a + b)$

According to the problem, this sum is equal to $k(4a + b)$. So we have:

$$\frac{4a + b}{16} = k(4a + b)$$

Now, dividing both sides by $4a + b$ (assuming $4a + b \neq 0$):

$$k = \frac{1}{16}$$

Thus, the value of $k$ is:

$$\boxed{\frac{1}{16}}$$

Extended Knowledge Points

What is the Sum of Solutions in a Quadratic Equation?

The sum of the solutions for any quadratic equation of the form $ax^2 + bx + c = 0$ is given by $-\frac{b}{a}$, where $b$ is the coefficient of $x$ and $a$ is the coefficient of $x^2$.

Quadratic Equations in General Form

A quadratic equation is typically written as $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. The solutions of this equation can be found using various methods, including factoring, completing the square, or the quadratic formula.

Application of Constants in Quadratic Problems

In many quadratic problems, constants such as $k$ represent proportionality or other scaling factors. Solving for these constants often requires equating terms or simplifying expressions to find the desired relationship.

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Similar Questions:

Question 1:

$25x^2 - (10m + 5n)x + mn = 0$
In the given equation, $m$ and $n$ are positive constants. Find the value of $k$.

Answer:

The value of $k$ is $1/5$.

Explanation:

Step 1: Identify the Equation Type

The given equation is a quadratic equation:

$$25x^2 - (10m + 5n)x + mn = 0$$

For any quadratic equation $Ax^2 + Bx + C = 0$, the sum of the solutions can be found with $-\frac{B}{A}$.

Step 2: Define $A$ and $B$

Here, we identify:

• $A = 25$
• $B = -(10m + 5n)$

Step 3: Calculate the Sum of Solutions

Using the sum of solutions formula:

$$\text{Sum of solutions} = -\frac{B}{A} = -\frac{-(10m + 5n)}{25} = \frac{10m + 5n}{25}$$

Step 4: Simplify the Sum of Solutions

Simplify $\frac{10m + 5n}{25}$ as follows:

$$\frac{10m + 5n}{25} = \frac{5(2m + n)}{25} = \frac{2m + n}{5}$$

Thus, the sum of the solutions is $\frac{2m + n}{5}$.

Step 5: Equate and Solve for $k$

According to the problem, the sum of the solutions is also $k(2m + n)$. Therefore:

$$\frac{2m + n}{5} = k(2m + n)$$

Step 6: Isolate $k$

Dividing both sides by $2m + n$ (assuming $2m + n \neq 0$):

$$k = \frac{1}{5}$$

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Question 2:

$9x^2 - (6p + 3q)x + pq = 0$
In the given equation, $p$ and $q$ are positive constants. Find the value of $k$.

Answer:

The value of $k$ is $1/3$.

Explanation:

Step 1: Identify the Equation Type

The given equation is a quadratic equation:

$$9x^2 - (6p + 3q)x + pq = 0$$

For any quadratic equation $Ax^2 + Bx + C = 0$, the sum of the solutions can be found with $-\frac{B}{A}$.

Step 2: Define $A$ and $B$

Here, we identify:

• $A = 9$
• $B = -(6p + 3q)$

Step 3: Calculate the Sum of Solutions

Using the sum of solutions formula:

$$\text{Sum of solutions} = -\frac{B}{A} = -\frac{-(6p + 3q)}{9} = \frac{6p + 3q}{9}$$

Step 4: Simplify the Sum of Solutions

Simplify $\frac{6p + 3q}{9}$ as follows:

$$\frac{6p + 3q}{9} = \frac{3(2p + q)}{9} = \frac{2p + q}{3}$$

Thus, the sum of the solutions is $\frac{2p + q}{3}$.

Step 5: Equate and Solve for $k$

According to the problem, the sum of the solutions is also $k(2p + q)$. Therefore:

$$\frac{2p + q}{3} = k(2p + q)$$

Step 6: Isolate $k$

Dividing both sides by $2p + q$ (assuming $2p + q \neq 0$):

$$k = \frac{1}{3}$$

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Question 3:

$16x^2 - (8r + 4s)x + rs = 0$
In the given equation, $r$ and $s$ are positive constants. Find the value of $k$.

Answer:

The value of $k$ is $1/4$.

Explanation:

Step 1: Identify the Equation Type

The given equation is a quadratic equation:

$$16x^2 - (8r + 4s)x + rs = 0$$

For any quadratic equation $Ax^2 + Bx + C = 0$, the sum of the solutions can be found with $-\frac{B}{A}$.

Step 2: Define $A$ and $B$

Here, we identify:

• $A = 16$
• $B = -(8r + 4s)$

Step 3: Calculate the Sum of Solutions

Using the sum of solutions formula:

$$\text{Sum of solutions} = -\frac{B}{A} = -\frac{-(8r + 4s)}{16} = \frac{8r + 4s}{16}$$

Step 4: Simplify the Sum of Solutions

Simplify $\frac{8r + 4s}{16}$ as follows:

$$\frac{8r + 4s}{16} = \frac{4(2r + s)}{16} = \frac{2r + s}{4}$$

Thus, the sum of the solutions is $\frac{2r + s}{4}$.

Step 5: Equate and Solve for $k$

According to the problem, the sum of the solutions is also $k(2r + s)$. Therefore:

$$\frac{2r + s}{4} = k(2r + s)$$

Step 6: Isolate $k$

Dividing both sides by $2r + s$ (assuming $2r + s \neq 0$):

$$k = \frac{1}{4}$$

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