Problem 102 - College Algebra

Question

Will help you prepare for the material covered in the next section. Solve by completing the square: \(y^{2}-6 y-4=0\)

Step-by-Step Solution

Answer

The solutions to the given quadratic equation are \(y=3+\sqrt{13}\) and \(y=3-\sqrt{13}\).

Step 1: Identify the coefficients

The coefficients are \(a=1\), \(b=-6\), and \(c=-4\).

Step 2: Rewrite the equation

Rewrite the equation into a form with a perfect square: \(y^{2}-6y+(-6/2)^{2}=4+(-6/2)^{2}\) which simplifies to \(y^{2}-6y+9=4+9\).

Step 3: Complete the square

This will result in a perfect square. So, the equation becomes \((y-3)^{2}=13\). The square of half of the coefficient of \(y\) which is 3, is subtracted and added to make a perfect square.

Step 4: Take square root on both sides

Computing square root on both sides gives \(y-3=\pm\sqrt{13}\). This is done because every positive real number has two square roots, one positive and the other negative.

Step 5: Solve for y

Finally, solve for \(y\) to get the two roots of the equation: \(y=3\pm\sqrt{13}\). This is done by transferring the 3 which is subtracted to the other side of the equation as addition.