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Zorluk: OrtaNonlinear Systems of Equations

Consider the system of equations below:

y=2x2+kx+9y=−2x+1\begin{aligned} y &= 2x^2 + kx + 9 \\ y &= -2x + 1 \end{aligned}

In the system of equations above, kk is a positive constant. If the system has exactly one real solution, what is the value of kk?

Cevap: 6

Cevap

The value of the positive constant kk is 6.
By substituting the linear equation into the quadratic equation, we obtain the equation 2x2+(k+2)x+8=02x^2 + (k+2)x + 8 = 0. For the system to have exactly one real solution, the discriminant of this quadratic equation, Δ=b2−4ac\Delta = b^2 - 4ac, must be equal to zero. Substituting a=2a = 2, b=k+2b = k+2, and c=8c = 8 into the discriminant formula gives (k+2)2−4(2)(8)=0(k+2)^2 - 4(2)(8) = 0, which simplifies to (k+2)2=64(k+2)^2 = 64. Solving for kk gives k+2=8k+2 = 8 or k+2=−8k+2 = -8, resulting in k=6k = 6 or k=−10k = -10. Since kk is specified to be a positive constant, the value of kk must be 6.

Adım Adım Çözüm

1
Substitute the expression for yy from the linear equation into the quadratic equation.
2x2+kx+9=−2x+12x^2 + kx + 9 = -2x + 1
Setting the two expressions for yy equal to each other allows us to solve for the xx-coordinates of the intersection points.
2
Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
2x2+(k+2)x+8=02x^2 + (k+2)x + 8 = 0
Grouping the terms by power of xx lets us identify the coefficients a=2a = 2, b=k+2b = k+2, and c=8c = 8.
3
Set the discriminant of the quadratic equation to zero.
(k+2)2−4(2)(8)=0(k+2)^2 - 4(2)(8) = 0, which simplifies to (k+2)2−64=0(k+2)^2 - 64 = 0
A quadratic equation has exactly one real solution if and only if its discriminant, b2−4acb^2 - 4ac, is equal to zero.
4
Solve for the positive constant kk.
k=6k = 6
Solving (k+2)2=64(k+2)^2 = 64 yields k+2=8k+2 = 8 or k+2=−8k+2 = -8, which means k=6k = 6 or k=−10k = -10. Since kk must be a positive constant, we select 6.

Anahtar Kavram

Determining the number of solutions of a quadratic-linear system using the discriminant.

Alternatif Yöntem

Alternatively, the quadratic equation 2x2+(k+2)x+8=02x^2 + (k+2)x + 8 = 0 has exactly one real solution when it is a perfect square trinomial. Factoring out the leading coefficient gives 2(x2+k+22x+4)=02(x^2 + \frac{k+2}{2}x + 4) = 0. For the term inside the parentheses to be a perfect square trinomial of the form x2+2dx+d2x^2 + 2dx + d^2, we must have d2=4d^2 = 4, which means d=±2d = \pm 2. Thus, the middle coefficient k+22\frac{k+2}{2} must equal 2d=±42d = \pm 4. Since kk is positive, we set k+22=4\frac{k+2}{2} = 4, which solves to k=6k = 6.
Tahmini Süre:1m 30s
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Konular

Nonlinear Systems of Equations

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