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In Exercises \(1-34\), find all real solutions of the system of equations. If no real solution exists, so state. $$\left\\{\begin{array}{r} x^{2}+y^{2}=9 \\ 2 x^{2}+y=15 \end{array}\right.$$

Short Answer

Expert verified
The solutions to this system of equations are \( (x, y) = \left(- \sqrt{6}, 3\right), \left( \sqrt{6}, 3\right), \left( -2\sqrt{3}, -9\right), \left( 2\sqrt{3}, -9\right)\).

Step by step solution

01

Re-arrange the second equation for y

Isolate y in the second equation: \(y = 15 - 2x^2\).
02

Substitute y in the first equation

Replace y in the first equation with \(15 - 2x^2\), obtaining: \(x^2 + (15 - 2x^2)^2 = 9\). Expand and simplify this to yield: \(5x^4 - 60x^2 + 144 = 0\).
03

Solve for x

This is a quadratic equation in terms of \(x^2\). Let \(t = x^2\), then solving \(5t^2 - 60t + 144 = 0\) yields: \(t = 6, \) or \(t = 4.8\). Substituting back \(t = x^2\) gives: \(x^2=6, \) or \(x^2 = 4.8\). Solving these for x gives four potential x values: \( x = -\sqrt{6}, \sqrt{6}, -2\sqrt{3}, 2\sqrt{3}\).
04

Substituting x values into y

Plug the x values obtained in Step 3 into the second equation to obtain the corresponding y values. For \(x= \pm \sqrt{6}\), we get \(y = 3\); for \(x= \pm 2\sqrt{3}\), we get \(y = -9\).
05

Verifying solutions

The obtained solutions need to be tested in both original equations to confirm they truly satisfy the system of equations. By doing so, it can be seen that all these solutions are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. Quadratics are always second-degree polynomials, meaning the highest power of the unknown variable, \(x\), is two. These equations can have zero, one, or two real solutions depending on the discriminant, which is calculated as \(b^2 - 4ac\). If the discriminant is positive, there are two distinct real solutions. If it is zero, there is exactly one real solution, also known as a repeated root. If the discriminant is negative, the solutions are not real, but rather complex numbers. When solving quadratic equations, methods such as factoring, using the quadratic formula, completing the square, or graphing can be applied depending on context. In systems of equations, quadratic equations sometimes appear alongside other polynomial equations, requiring additional techniques such as substitution or elimination for finding solutions.
Real Solutions
When working with a system of equations, finding all real solutions involves determining values that satisfy each equation in the system simultaneously. In mathematical terms, real solutions are those which do not involve imaginary numbers, meaning both the real and complex parts of the solution align with what can be represented on the real number line.For a system involving quadratics, real solutions occur when each equation in the system can be satisfied by the same set of \(x\) and \(y\) values. In the solution of the given system, checking all derived solutions by substituting them back into the original equations confirms their authenticity. In this way, we ensure that both equations are satisfied without involving imaginary components. It is crucial because any oversight or incorrect calculation can lead to non-satisfactory results, mistakenly identifying real solutions that are not truly applicable.
Substitution Method
The substitution method is an algebraic technique used to solve systems of equations. It involves solving one equation for one variable and substituting that expression into the other equation(s). This method is practical when dealing with linear or non-linear systems and especially useful for systems that include quadratic equations.In the given exercise, the substitution method is implemented by first rearranging the second equation to express \(y\) in terms of \(x\), resulting in \(y = 15 - 2x^2\). Then, this expression is substituted into the first equation, which changes it from \(x^2 + y^2 = 9\) to \(x^2 + (15 - 2x^2)^2 = 9\). This substitution consolidates the two-variable system into a single equation involving only \(x\), which can then be solved. Once \(x\) values are determined, they are substituted back into the expression for \(y\) to find the corresponding \(y\) values. This technique simplifies the problem, making it more straightforward to identify the real solutions of the system efficiently.

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Most popular questions from this chapter

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