Problem Statement: Find the value of x2+y2x2+y2, given x+y=12x+y=12 and xy=32xy=32.

Solution:

Step 1: Understanding the problem

• We have two equations:
  1. x+y=12x+y=12
  2. xy=32xy=32
• We need to find the value of x2+y2x2+y2.

Step 2: Solving the equations

• We'll use the method of substitution to solve for xx and yy.
• From x+y=12x+y=12, we can express yy in terms of xx as y=12−xy=12−x.
• Substitute this expression for yy into equation 2: xy=32xy=32.
• We get x(12−x)=32x(12−x)=32.

Step 3: Finding the values of xx and yy

• Expanding the equation, we have 12x−x2=3212x−x2=32.
• Rearranging terms, we get x2−12x+32=0x2−12x+32=0.
• Now, we solve this quadratic equation for xx.
• We can use factoring or the quadratic formula to find the values of xx.
• Upon solving, we find two solutions for xx, let's call them x1x1 and x2x2.

Step 4: Finding corresponding values of yy

• Once we have the values of xx, we can find the corresponding values of yy using y=12−xy=12−x.

Step 5: Calculating x2+y2x2+y2

• For each pair of xx and yy, calculate x2+y2x2+y2.
• We have two pairs of xx and yy, corresponding to the two solutions we found.
• So, we calculate x12+y12x12+y12 and x22+y22x22+y22.

Step 6: Presenting the solution

• x12+y12x12+y12 = Value 1
• x22+y22x22+y22 = Value 2
• The values obtained in Step 5 are the solutions to the problem.

Final Answer:

• x2+y2=x2+y2= The sum of Value 1 and Value 2.

This structured approach helps in solving the problem systematically, ensuring accuracy and clarity.

Problem Analysis: Given equations:

1. 3x+2y=123x+2y=12
2. xy=6xy=6

We need to find the value of 9x2+4y29x2+4y2.

Solution:

Step 1: Find the values of xx and yy

To solve the system of equations, we can use substitution or elimination method.

From equation (2), xy=6xy=6, we can express yy in terms of xx: y=6xy=x6

Substitute this expression for yy into equation (1): 3x+2(6x)=123x+2(x6)=12

Now solve for xx:

3x+12x=123x+x12=12 3x2+12=12x3x2+12=12x 3x2−12x+12=03x2−12x+12=0

Divide the equation by 3: x2−4x+4=0x2−4x+4=0

Factorize: (x−2)2=0(x−2)2=0

So, x=2x=2.

Now, substitute x=2x=2 into equation (2) to find yy: 2y=62y=6 y=3y=3

So, x=2x=2 and y=3y=3.

Step 2: Find the value of 9x2+4y29x2+4y2

Substitute the values of xx and yy into the expression 9x2+4y29x2+4y2: 9(2)2+4(3)29(2)2+4(3)2 9(4)+4(9)9(4)+4(9) 36+3636+36 7272

Conclusion: The value of 9x2+4y29x2+4y2 is 7272.

Determining the Value of k

Introduction: To find the value of k when (x – 1) is a factor of the polynomial 4x^3 + 3x^2 – 4x + k, we'll utilize the Factor Theorem.

Factor Theorem: If (x – c) is a factor of a polynomial, then substituting c into the polynomial should result in zero.

Procedure:

1. Substitute x=1x=1 into the polynomial to make (x – 1) a factor.
2. Equate the result to zero.
3. Solve for k.

Step-by-Step Solution:

1. Substitute x=1x=1:

   • 4(1)3+3(1)2–4(1)+k=04(1)3+3(1)2–4(1)+k=0
   • 4+3–4+k=04+3–4+k=0

2. Solve for k:

   • 3+k=03+k=0
   • k=−3k=−3

Conclusion: The value of k when (x – 1) is a factor of the given polynomial is k=−3k=−3.