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Zorluk: ZorSimultaneous Linear and Quadratic Equations

Given the simultaneous equations 2x+y=52x + y = 5 and x2+xyy2=5x^2 + xy - y^2 = -5, let (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) be the real solution pairs. What is the value of x1x2+y1y2x_1 x_2 + y_1 y_2?

  1. 5-5Cevap
  2. B
    1313
  3. C
    55
  4. D
    9595

Cevap

-5
Substituting y=52xy = 5 - 2x into x2+xyy2=5x^2 + xy - y^2 = -5 yields 5x2+25x20=0-5x^2 + 25x - 20 = 0, which simplifies to x25x+4=0x^2 - 5x + 4 = 0. The roots are x1=1x_1 = 1 and x2=4x_2 = 4, giving corresponding yy-values y1=3y_1 = 3 and y2=3y_2 = -3. Evaluating x1x2+y1y2x_1 x_2 + y_1 y_2 gives (1)(4)+(3)(3)=49=5(1)(4) + (3)(-3) = 4 - 9 = -5.

Adım Adım Çözüm

1
Express yy in terms of xx using the linear equation.
y=52xy = 5 - 2x
Substitution method requires isolating one variable from the linear equation.
2
Substitute y=52xy = 5 - 2x into the quadratic equation x2+xyy2=5x^2 + xy - y^2 = -5.
x2+x(52x)(52x)2=5x^2 + x(5 - 2x) - (5 - 2x)^2 = -5
Form a single quadratic equation in terms of xx.
3
Expand and simplify the quadratic equation.
x2+5x2x2(2520x+4x2)=5    5x2+25x20=0    x25x+4=0x^2 + 5x - 2x^2 - (25 - 20x + 4x^2) = -5 \implies -5x^2 + 25x - 20 = 0 \implies x^2 - 5x + 4 = 0
Convert the equation to standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Factorize the quadratic equation to find the values of xx.
(x1)(x4)=0    x1=1,x2=4(x - 1)(x - 4) = 0 \implies x_1 = 1, x_2 = 4
Obtain the two roots for xx.
5
Find the corresponding yy-values using y=52xy = 5 - 2x.
For x1=1x_1 = 1: y1=52(1)=3y_1 = 5 - 2(1) = 3. For x2=4x_2 = 4: y2=52(4)=3y_2 = 5 - 2(4) = -3. Solution pairs are (1,3)(1, 3) and (4,3)(4, -3).
Calculate corresponding coordinate values for each root.
6
Compute the required expression x1x2+y1y2x_1 x_2 + y_1 y_2.
x1x2+y1y2=(1)(4)+(3)(3)=49=5x_1 x_2 + y_1 y_2 = (1)(4) + (3)(-3) = 4 - 9 = -5
Perform the final calculation requested in the stem.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations
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