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Zorluk: ZorNonlinear Systems of Equations

A system of two equations is given:

y=x2+kx+4y = x^2 + kx + 4
y=4x−ky = 4x - k

In these equations, kk represents a positive constant. If the system has a single real solution, what is the value of kk?

Cevap: 12

Cevap

12
Setting the two equations equal yields the quadratic equation x2+(k−4)x+(k+4)=0x^2 + (k - 4)x + (k + 4) = 0. For the system to have exactly one real solution, the discriminant of this quadratic equation must be equal to 00. The discriminant is (k−4)2−4(1)(k+4)=k2−12k(k - 4)^2 - 4(1)(k + 4) = k^2 - 12k. Solving k2−12k=0k^2 - 12k = 0 yields k=0k = 0 or k=12k = 12. Since kk must be positive, the value of kk is 1212.

Adım Adım Çözüm

1
Set the two equations equal to each other to form a single quadratic equation in terms of xx.
x2+(k−4)x+(k+4)=0x^2 + (k - 4)x + (k + 4) = 0
Equating the expressions for yy allows us to find the xx-coordinates where the graphs of the two equations intersect.
2
Set the discriminant of the quadratic equation to zero.
(k−4)2−4(1)(k+4)=0(k - 4)^2 - 4(1)(k + 4) = 0
A system of equations consisting of a line and a parabola has a single real solution if and only if the line is tangent to the parabola, which corresponds to a quadratic equation with a discriminant of zero.
3
Expand and simplify the equation for kk.
k2−12k=0k^2 - 12k = 0
Expanding (k−4)2(k - 4)^2 yields k2−8k+16k^2 - 8k + 16, and distributing −4-4 yields −4k−16-4k - 16. Combining like terms simplifies the relation.
4
Solve for kk and apply the constraint that k>0k > 0.
k=12k = 12
Factoring k(k−12)=0k(k - 12) = 0 gives k=0k = 0 or k=12k = 12. Since kk is specified to be positive, k=12k = 12 is the correct value.

Anahtar Kavram

Solving nonlinear systems of equations by substitution and using the discriminant to find conditions for a single real solution.
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Konular

Nonlinear Systems of Equations

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