Solving Quadratic Equations by Substitution

To finish our problem, we now need to find what our x equals. To do this, we go back to our substitution, u = x^2. We plug this in for the us we found. We get x^2 = 1 and x^2 = 2. Solving for x, we get x = +/- 1 and x = +/- sqrt(2).

In this case, we have a total of four answers. This is because our super equation has a degree of 4. A good check to make sure we have all the answers that we need is to look at the degree of our super equation, our quadratic equation in disguise. Whatever degree that equation is, is the number of solutions that we will find. For our problem, we end up with x = 1, x = -1, x = sqrt(2), and x = - sqrt(2).

We can make the substitution u = x - 2 since x - 2 is the disguise of this super equation. Our equation then becomes the quadratic u^2 - 4u - 12. Factoring, we get (u - 6)(u + 2). Solving for u, we get u = 6 and u = -2. Substituting our u = x - 2 back in, we get x - 2 = 6 and x - 2 = -2. Solving for x now, we get x = 8 and x = 0. And we are done!

Quadratic equations are polynomials of degree 2. Sometimes, we will see our quadratic equations disguised as super equations such as x^4 + 3x^2 + 2. We know these are disguised quadratics because by making one substitution, we can turn the super equation into a normal quadratic equation that we can easily solve. After we make the substitution, we solve the quadratic equation as we normally do using the skills we have already learned, such as factoring. Then we replace our substitution to find our final answer in the variable we began with.

• Define quadratic equations
• Explain how to solve super equations/disguised quadratics by substitution