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Difficulty: HardNonlinear Systems of Equations

In the system of equations below, kk is a positive constant.

x−y=kx - y = k
x2−3xy+y2=5x^2 - 3xy + y^2 = 5

If the system has exactly one real solution (x,y)(x, y), what is the value of kk?

Answer: 2

Answer

The value of kk is 2.
Substituting y=x−ky = x - k into the second equation yields −x2+kx+k2−5=0-x^2 + kx + k^2 - 5 = 0, which can be rewritten in standard form as x2−kx+(5−k2)=0x^2 - kx + (5 - k^2) = 0. For this quadratic equation to have exactly one real solution, its discriminant must equal 0: b2−4ac=(−k)2−4(1)(5−k2)=5k2−20=0b^2 - 4ac = (-k)^2 - 4(1)(5 - k^2) = 5k^2 - 20 = 0. Solving 5k2−20=05k^2 - 20 = 0 gives k2=4k^2 = 4, and since kk must be positive, k=2k = 2.

Step-by-Step Solution

1
Rearrange the first equation to express yy in terms of xx.
y=x−ky = x - k
This allows for substitution into the second equation to eliminate yy.
2
Substitute y=x−ky = x - k into the second equation and expand.
x2−3x(x−k)+(x−k)2=5  ⟹  −x2+kx+k2−5=0x^2 - 3x(x - k) + (x - k)^2 = 5 \implies -x^2 + kx + k^2 - 5 = 0
To create a single quadratic equation in terms of xx.
3
Multiply by −1-1 to write the quadratic equation in standard form ax2+bx+c=0ax^2 + bx + c = 0.
x2−kx+(5−k2)=0x^2 - kx + (5 - k^2) = 0
Standard form makes it easier to identify the coefficients a=1a = 1, b=−kb = -k, and c=5−k2c = 5 - k^2.
4
Set the discriminant b2−4acb^2 - 4ac equal to 0.
(−k)2−4(1)(5−k2)=0(-k)^2 - 4(1)(5 - k^2) = 0
A quadratic equation has exactly one real solution if and only if its discriminant is zero.
5
Simplify the discriminant equation and solve for kk.
5k2−20=0  ⟹  k2=4  ⟹  k=25k^2 - 20 = 0 \implies k^2 = 4 \implies k = 2 (since kk must be positive)
To find the positive constant kk that satisfies the condition.

Key Concept

Determining the number of solutions to a nonlinear system by substituting and setting the discriminant of the resulting quadratic equation to zero.

Alternative Method

Alternatively, one can rewrite the second equation by grouping: x2−3xy+y2=(x−y)2−xy=5x^2 - 3xy + y^2 = (x - y)^2 - xy = 5. Since x−y=kx - y = k, we have k2−xy=5k^2 - xy = 5, so xy=k2−5xy = k^2 - 5. We now have a system of x−y=kx - y = k and xy=k2−5xy = k^2 - 5. Substituting y=x−ky = x - k gives x(x−k)=k2−5x(x - k) = k^2 - 5, leading to x2−kx+(5−k2)=0x^2 - kx + (5 - k^2) = 0, which can be solved using the discriminant as shown in the primary method.
Estimated Time:2m 30s
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Topics

Nonlinear Systems of Equations

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