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**Equations in Quadratic Form**

The "u" Substitution Method

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**Set each factor = 0 and solve**

Before we solve the above equation, let's solve a quadratic equation that we know how to solve. Factor Set each factor = 0 and solve Let's use this to solve the original equation by letting u = x2.

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**Set each factor = 0 and solve**

If u = x2 then square both sides and get u2 = x4. Substitute u and u2 for x2 and x4. Factor Set each factor = 0 and solve Now that we've solved for u we have to re-substitute to get x back. Remember u = x2 so let's substitute. Solve for x by square-rooting both sides and don't forget the

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**Factor & set each factor = 0 and solve**

You can determine if an equation is of quadratic form where you can use the "u" substitution method if you call the middle variable and power u and then square it and get the first term's variable and power. So let u = z1/4 and get u2 = z1/2. Substitute u and u2 for z1/4 and z1/2. Factor & set each factor = 0 and solve Solve for z by raising both sides to the 4th power

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**Factor & set each factor = 0 and solve**

Let's try one more. Call the middle variable u and then square it to see if you get the first term's variable. So let u = x3 and get u2 = x6. Substitute u and u2 for x3 and x6. Factor & set each factor = 0 and solve Solve for x by taking the cube root of both sides

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Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar

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The standard form of the equation of a circle with its center at the origin is Notice that both the x and y terms are squared. Linear equations don’t.

The standard form of the equation of a circle with its center at the origin is Notice that both the x and y terms are squared. Linear equations don’t.

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