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Chapter 13: Problem 39

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} y=(x+3)^{2} \\ x+2 y=-2 \end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are (-4, 1) and (-5/2, 1/4).

Step by step solution

01

- Substitution

Now that we have \(y = (x + 3)^2\), we're going to substitute this into our second equation where we have 'y'. So, our second equation \(x + 2y = -2\) becomes \(x + 2(x+3)^2 = -2\)
02

- Simplify the equation

Expand and simplify the equation to form a quadratic equation. Expansion gives the equation as \(x + 2(x^2 + 6x + 9) = -2\) which on further simplification we get \(2x^2 + 12x + 18 + x + 2 = 0\), which gives \(2x^2 + 13x + 20 = 0\).
03

- Solve the quadratic equation

We can solve the quadratic equation \(2x^2 + 13x + 20 = 0\) by factorization method (because the given quadratic expression can be factored). The factors of the quadratic equation are \(2x^2 + 8x+ 5x + 20 = 0\). Factoring by grouping gives us \(2x(x+4)+5(x+4) = 0\) which further gives \((x+4)(2x+5) = 0\).
04

- Calculate x values

Setting each group equal to zero gives the roots of the equation. This implies x=-4 or x=-5/2. These are the two possible x values that satisfy the system of equations.
05

- Calculate y values

Now plug these x values back into the first equation to find the corresponding y values. For \(x=-4\), we get \(y = (-4 + 3)^2 = (-1)^2 = 1\) and for \(x=-5/2\), we get \(y = (-5/2+3)^2 = (-1/2)^2 = 1/4\). So the solutions are (-4, 1) and (-5/2, 1/4).

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Most popular questions from this chapter

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$x^{2}+4 y^{2}=16$$

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. I'm graphing an equation that contains neither an \(x^{2}\) -term nor a \(y^{2}\) -term, so the graph cannot be a conic section.

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $$4 x^{2}=36+y^{2}$$

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y=-x^{2}-4 x+4$$

Use the vertex and intercepts to sketch the graph of each equation. If needed, find additional points on the parabola by choosing values of y on each side of the axis of symmetry. $$x=-(y-5)^{2}+4$$
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