This method will solve any quadratic equation and can be used when easier methods don’t work. It is also, by the way, the basis for deriving the quadratic formula that every high school student should have memorized.
Let’s learn the method and then, for those interested, we can learn the theory.
Example 1. Solve \(x^{2}+6x-10=0\)
Step 1. Make the coefficient of the \(x^{2}\) term ‘1’. (Oh goodie, that’s already the case)
Step 2. Make a perfect square. In this case, add ’19’ to both sides: \(x^{2}+6x-9+19=+19\) which gives
\(x^{2}+6x+9=19\)
Step 3. Factor the perfect square on the left…
\((x+3)^{2}=19\)
Step 4. Take the square root of both sides:
\(\sqrt{(x+3)^{2}}=\sqrt{19}\)
\(x+3=\sqrt{19}\)
Step 5. Solve for x. (gives two roots)
\(x=-3+\sqrt{19}\)
\(x=-3-\sqrt{19}\)
Example 2. Solve \(3x^{2}-8x+10=0\)
Step 1. Make the coefficient of the \(3x^{2}\) term ‘1’ by dividing by ‘3’.
\(x^{2}-\frac{8}{3}x+\frac{10}{3}=0\)
Step 2. Make a perfect square. In this case, we need to make the third term equal to half of the second term squared by subtracting \((\frac{8}{6})^2\) from both sides and moving the \((\frac{10}{3})^2\) term to the right side.
\(x^{2}-\frac{8}{3}x-(\frac{8}{6})^2=-(\frac{8}{6})^2-\frac{10}{3}\) giving:
\(x^{2}-\frac{8}{3}x-(\frac{8}{6})^2=-(\frac{64}{36})-\frac{130}{36}\) which then simplifies to:
\(x^{2}-\frac{8}{3}x-(\frac{8}{6})^2=-(\frac{194}{36})\)
Step 3. Now factor the perfect square on the left…
\((x-\frac{4}{3})^{2}=-(\frac{{194}}{36})\)
Step 4. Then take the square root of both sides:
\(\sqrt{(x-\frac{4}{3})^{2}}=\sqrt{-\frac{194}{36}}\)
\(x-\frac{4}{3}=\sqrt{-1}\sqrt{\frac{{194}}{36}}\)
Step 5. Solve for x. (note we end up with complex roots)
\(x=\frac{4}{3}+i\sqrt{\frac{194}{36}}\)
\(x=\frac{4}{3}-i\sqrt{\frac{194}{36}}\)
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