Give all solutions for the nonlinear system of equations, including those with real and complex components.

\[ y = x^2 + 6x \]

\[ 4x - y = -24 \]

y=x^2+6x Apply the value of y into the equation below: 4x-y=-24 4x-(x^2+6x) = -24 4x-x^2-6x = -24 x^2+2x-24=0 (x+6)(x-4)=0 Therefore x can be -6 or 4 If x = -6 Then y = (-6)^2+6(-6) = 36 -36=0 If x = 4 Then y = (4)^2 + 6(4) = 16 + 24 = 40

Answered by leonghw | 2024-06-10

\left\{\begin{array}{ccc}y=x^2+6x\\4x-y=-24\end{array}\right\\\\substitute:\\\\4x-(x^2+6x)=-24\\\\4x-x^2-6x+24=0\\\\-x^2-2x+24=0\\\\a=-1;\ b=-2;\ c=24\\\\\Delta=b^2-4ac
Δ = ( − 2 ) 2 − 4 ⋅ ( − 1 ) ⋅ 24 = 4 + 96 = 100 x 1 ​ = 2 a − b − Δ ​ ​ ; x 2 ​ = 2 a − b + Δ ​ ​ Δ ​ = 100 ​ = 10 x 1 ​ = 2 ⋅ ( − 1 ) 2 − 10 ​ = − 2 − 8 ​ = 4 ; x 2 ​ = 2 ⋅ ( − 1 ) 2 + 10 ​ = − 2 12 ​ = − 6 y 1 ​ = 4 2 + 6 ⋅ 4 = 16 + 24 = 40 ; y 2 ​ = ( − 6 ) 2 + 6 ⋅ ( − 6 ) = 36 − 36 = 0 A n s w er : x = 4 an d y = 40 or x = − 6 an d y = 0

Answered by Anonymous | 2024-06-10

The solutions to the nonlinear system of equations are (4, 40) and (-6, 0). The first equation is a quadratic equation, and the second is a linear equation that both yield real solutions. Thus, there are no complex components in this case.
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Answered by Anonymous | 2024-12-17