Question

Solve the system by the method of substitution. $$\left\\{\begin{array}{l}y=x^{4}-2 x^{2}+1 \\ y=1-x^{2}\end{array}\right.$$

Short answer

The solutions to the system of equations are (0, 1), (\sqrt{3}, -2), and (-\sqrt{3}, -2)

Step-by-step solution

Isolate y in both equations

We see that y is already isolated in both the equations. So we use the substitution method.

Substitute y from second equation into first

The idea is to put \(1 - x^2\) where we see \(y\) in the first equation. Hence, the first equation becomes: \(1 - x^2 = x^4 - 2x^2 + 1\)

Rearrange the equation

Rearrange the terms to form a common quadratic equation: \(0 = x^4 - 2x^2 - x^2 + 1 - 1\),which simplifies to \(0 = x^4 - 3x^2\)

Solve the resulting equation

Factor out an \(x^2\):\(0 = x^2(x^2 - 3)\)This yields two potential solutions for x: \(x = 0\) or \(x = \sqrt{3}, -\sqrt{3}\}

Substitute x into second equation to find y

Substitute these calculated x values into our second equation to calculate y. Substituting \( x = 0 \) we get \(y = 1 - 0 = 1\). Substituting \( x = \sqrt{3} \) or \( x = -\sqrt{3} \) we get \( y = 1 - 3 = -2 \).

Key concepts

Substitution Method in Algebra

The substitution method is a powerful tool in algebra for solving systems of equations, where one variable is replaced with an expression containing the other variable. To use this method, we must have one of the variables isolated in at least one of the equations, which provides us with a substitution equation. In the given exercise, both equations have already isolated the variable y, making it straightforward to apply substitution.

Here's a simplified approach to understand substitution:

• Isolate y (or x) in one equation if it’s not already isolated.
• Substitute the expression for the isolated variable into the other equation.
• Solve the new equation for the remaining variable.
• Back-substitute the found value into any of the original equations to find the value of the other variable.

This sequential method converts a system of equations into a single equation in one variable, which can generally be solved more easily.

Solving Quadratic Equations

Quadratic equations appear frequently in algebra and are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). To solve such equations, we must find the value(s) of x that make the equation true. There are several methods to solve quadratic equations: factoring, using the quadratic formula, completing the square, and graphing.

When using the substitution method, we often simplify the system to a quadratic equation, as in our example exercise. Here, we ended up with \(x^4 - 3x^2 = 0\), which we treated as a quadratic by factoring \(x^2\) out. This factorization converted the higher-degree equation into a solvable quadratic form, allowing us to find the solutions for x. It's important to recognize such opportunities to simplify equations to quadratics as they present a clear path to finding the solutions.

Factoring Quadratic Expressions

Factoring is a method of rewriting an expression as a product of its factors. For quadratics, this often involves finding two binomials that, when multiplied together, give the original quadratic expression. In the context of our exercise, we had to factor \(x^4 - 3x^2 = 0\), which isn't a quadratic in its initial form. However, by recognizing \(x^2\) as a common term, we essentially reformatted the equation into \(x^2(x^2 - 3) = 0\), thus simplifying the complex equation into factors that resemble a quadratic.

Factoring is most straightforward when the quadratic is a simple trinomial, but there are techniques for factoring more complex quadratics or those that require substitution to recognize their quadratic nature. Once factored, we set each factor equal to zero to solve for x, because if the product of two factors is zero, at least one of the factors must be zero. This fundamental principle of algebra underpins the factoring technique to solving quadratic equations.