/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Solve each system. $$\begin{al... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 24

Solve each system. $$\begin{aligned} &y-x=1\\\ &4 y^{2}-16 x^{2}=64 \end{aligned}$$

Short Answer

Expert verified
The given system of equations \[ \begin{cases} y-x=1\\ 4y^2-16(x-1)^2=64 \end{cases} \] has no real solutions, because after substituting \(x=y-1\) into the second equation, simplifying, and attempting to solve the resulting quadratic equation, we find a negative discriminant, which means there are no real solutions for y.

Step by step solution

01

Solve the first equation for x or y

The first equation is \(y-x=1\). Let's solve for x: \(x = y-1\)
02

Substitute the expression for x into the second equation

Now substitute the expression we found for x into the second equation: \(4 y^2 - 16(y - 1)^2=64\)
03

Simplify and solve for y

First, expand the second equation and simplify: \[4y^2 - 16(y^2 - 2y + 1) = 64\] \[4y^2 - 16y^2 + 32y - 16 = 64\] Now, collect like terms and solve for y: \[-12y^2 + 32y - 80 = 0\] Divide through by -4: \[3y^2 - 8y + 20 = 0\]
04

Solve the quadratic equation for y

We can use the quadratic formula to solve for y: \[y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] For our equation, a=3, b=-8, and c=20. \[y=\frac{8\pm\sqrt{(-8)^2-4(3)(20)}}{2(3)}\] Now calculate the discriminant: \((-8)^2-4(3)(20)=64-240=-176\) Since the discriminant is negative, the quadratic equation has no real solutions.
05

Conclusion

Since there are no real solutions for y in the quadratic equation, the system of equations has no real solutions. The given system of equations cannot be solved for real values of x and y.

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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 equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. These equations involve one unknown variable, usually \(x\), and the highest exponent of the variable is 2. The graph of a quadratic equation is a parabola.
  • The vertex of the parabola can be a maximum or minimum point.
  • Quadratic equations can have zero, one, or two solutions depending on the discriminant (which is related to the roots).
Quadratic equations can be solved using different methods, such as factoring, completing the square, or using the quadratic formula. For systems of equations involving quadratic expressions, solving one equation for a variable and substituting it into the other can sometimes simplify the process.
Discriminant
The discriminant is part of the quadratic formula and helps determine the nature of the solutions of a quadratic equation. It is found using the expression \(b^2 - 4ac\) where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).
  • If the discriminant is positive, there are two distinct real solutions.
  • If it is zero, there is exactly one real solution (a repeated root).
  • If the discriminant is negative, there are no real solutions, but two complex solutions.
In the given problem, the discriminant \(-176\) is negative, indicating that the quadratic equation has no real solutions, leading to the conclusion that the system of equations cannot be solved for real numbers.
Substitution Method
The substitution method is a technique used for solving systems of equations, particularly when one equation can be easily solved for one of the variables. The steps include:
  • Solve one equation for one variable.
  • Substitute the expression for the solved variable into the other equation.
  • Simplify and solve the resulting equation.
This method works well when one of the equations is simple enough to solve for one variable. In the original exercise, we initially solved for \(x\) in the first linear equation and substituted it into the quadratic equation, which helped simplify our approach to finding potential solutions for \(y\).
No Real Solutions
When a system of equations, like the one in the problem, results in no real solutions, it means there are no points on a Cartesian plane that satisfy both equations simultaneously. The reasons can include:
  • For linear-quadratic systems, the quadratic graph (parabola) and the line may not intersect at all.
  • This typically occurs if the solutions involve imaginary or complex numbers, indicated by a negative discriminant in the quadratic equation.
Understanding that equations can have no real solutions is crucial in cases involving quadratic equations. It is important to recognize when the algebra leads to such outcomes, as this affects the graphical interpretation and practical applications of these equations.

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Most popular questions from this chapter

Solve the system of equations. $$\begin{aligned} x+y+z &=2 \\ 6 x-4 y+5 z &=31 \\ 5 x+2 y+2 z &=13 \end{aligned}$$

Use a graphing calculator to find the approximate solutions of the equation. $$\log _{3} x+7=4-\log _{5} x$$

Use a graphing calculator to find the approximate solutions of the equation. $$\log _{5}(x+7)-\log _{5}(2 x-3)=1$$

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{2}(x+20)-\log _{2}(x+2)=\log _{2} x$$

a) Find the vertex. b) Find the axis of symmetry. c) Determine whether there is a maximum or minimum value and find that value. $$g(x)=x^{2}-6$$
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