Problem 30
Question
Solve each system. $$\begin{array}{l} x^{2}+y^{2}=1 \\ y=x^{2}+1 \end{array}$$
Step-by-Step Solution
Verified Answer
The given system of equations represents a circle of radius 1 centered at the origin and a parabola opening upwards with a vertex at (0,1). By substituting the second equation into the first, we get \(x^2 + (x^2 + 1)^2 = 1\). Simplifying and factoring, we find that \(x^2(x^2 + 3) = 0\). The only real solution for x is \(x = 0\). Plugging this back into the second equation, we get \(y = (0)^2 + 1 = 1\). The only point of intersection for the given system is \((x, y) = (0, 1)\).
1Step 1: Substitute the second equation into the first equation
Replace y in the first equation with the expression for y from the second equation:
\(x^2 + (x^2 + 1)^2 = 1\)
2Step 2: Simplify the equation and solve for x
Expand and simplify the equation:
\[x^2 + (x^4 + 2x^2 + 1) = 1\]
\[x^4 + 3x^2 = 0\]
Now we can factor the equation to find the solutions for x:
\[x^2(x^2 + 3) = 0\]
Thus, x has two possible solutions: \(x = 0\) or \(x^2 = -3\) (which has no real solutions, so we'll discard that).
3Step 3: Solving for y
Plug the solution for x back into the second equation to find the corresponding value of y:
\[y = (0)^2 + 1 = 1\]
4Step 4: Identify the points of intersection
The only point of intersection for the given system is \((x, y) = (0, 1)\).
Key Concepts
Substitution MethodAlgebraic FactoringSolving Quadratic EquationsIntersection Points
Substitution Method
The substitution method is a powerful tool for solving systems of equations. It involves expressing one variable in terms of the other using one equation, then replacing that expression in the other equation. This approach simplifies the problem to a single equation with one variable.
For instance, in our exercise, the second equation is given as \( y = x^2 + 1 \). We can express \( y \) in terms of \( x \), then substitute \( x^2 + 1 \) in place of \( y \) in the first equation, \( x^2 + y^2 = 1 \).
This process converts the problem into a single equation: \( x^2 + (x^2 + 1)^2 = 1 \), which can then be solved for \( x \).
Substitution is often used when it's straightforward to solve one of the equations for a specific variable.
For instance, in our exercise, the second equation is given as \( y = x^2 + 1 \). We can express \( y \) in terms of \( x \), then substitute \( x^2 + 1 \) in place of \( y \) in the first equation, \( x^2 + y^2 = 1 \).
This process converts the problem into a single equation: \( x^2 + (x^2 + 1)^2 = 1 \), which can then be solved for \( x \).
Substitution is often used when it's straightforward to solve one of the equations for a specific variable.
Algebraic Factoring
Algebraic factoring is a fundamental technique for simplifying polynomial equations. Once a polynomial is in its factored form, it's easier to identify its roots.
In our example, the simplified equation \( x^4 + 3x^2 = 0 \) can be factored as \( x^2(x^2 + 3) = 0 \). The solution involves finding the values of \( x \) that satisfy this equation.
In our example, the simplified equation \( x^4 + 3x^2 = 0 \) can be factored as \( x^2(x^2 + 3) = 0 \). The solution involves finding the values of \( x \) that satisfy this equation.
- We see that either \( x^2 = 0 \) or \( x^2 + 3 = 0 \).
- \( x^2 = 0 \) gives us a real solution: \( x = 0 \).
- \( x^2 = -3 \) doesn't yield real solutions, as you cannot take a real square root of a negative number in basic algebra.
Solving Quadratic Equations
Solving quadratic equations is a key task in algebra. Quadratic equations are of the form \( ax^2 + bx + c = 0 \) and can often be solved by factoring, using the quadratic formula, or completing the square.
In our exercise, we simplified to \( x^4 + 3x^2 = 0 \) and then factored to \( x^2(x^2 + 3) = 0 \). The term \( x^2 = 0 \) is a quadratic form that can be directly solved, giving \( x = 0 \).
The other part, \( x^2 + 3 = 0 \), does not provide real solutions in basic algebra. If you were to solve it with complex numbers, you would find that \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \). However, since we are focusing on real numbers, these solutions are not considered in this context.
In our exercise, we simplified to \( x^4 + 3x^2 = 0 \) and then factored to \( x^2(x^2 + 3) = 0 \). The term \( x^2 = 0 \) is a quadratic form that can be directly solved, giving \( x = 0 \).
The other part, \( x^2 + 3 = 0 \), does not provide real solutions in basic algebra. If you were to solve it with complex numbers, you would find that \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \). However, since we are focusing on real numbers, these solutions are not considered in this context.
Intersection Points
Finding intersection points means identifying where two graphs meet. These points represent solutions that satisfy both equations in a system.
In our example, we solve the system to find the coordinates \((x, y)\) that satisfy both equations, \( x^2 + y^2 = 1 \) and \( y = x^2 + 1 \).
After solving and simplifying, we found that \( x = 0 \) is the only real solution. Plugging \( x = 0 \) back into \( y = x^2 + 1 \), we find \( y = 1 \).
Thus, the only intersection point is \((0, 1)\), where both equations intersect when plotted. Intersection point solutions are critical, as they denote the exact places where two different equations or graphs align.
In our example, we solve the system to find the coordinates \((x, y)\) that satisfy both equations, \( x^2 + y^2 = 1 \) and \( y = x^2 + 1 \).
After solving and simplifying, we found that \( x = 0 \) is the only real solution. Plugging \( x = 0 \) back into \( y = x^2 + 1 \), we find \( y = 1 \).
Thus, the only intersection point is \((0, 1)\), where both equations intersect when plotted. Intersection point solutions are critical, as they denote the exact places where two different equations or graphs align.
Other exercises in this chapter
Problem 30
Solve each system. $$\begin{aligned} &x^{2}+y^{2}=1\\\ &y=x^{2}+1 \end{aligned}$$
View solution Problem 30
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In New York City, orange juice, a raisin bagel, and a cup of coffee from Katie's Koffee Kart cost a total of \(\$ 8.15 .\) Katie posts a notice announcing that,
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Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{5} x=4$$
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