91Ó°ÊÓ

A quadratic equation is a second-order polynomial equation in a single variable, typically written in the form \[ ax^2 + bx + c = 0 \]where *a*, *b*, and *c* are constants. To solve these equations, we need to find the values of *x* that make the equation true. In our exercise, we started with the function \( g(x) = 2x^2 - 15x \) and set it equal to -7, forming the equation:
\[ 2a^2 - 15a + 7 = 0 \]
There are several methods to solve quadratic equations:

• Factoring
• Completing the square
• Using the quadratic formula

In our case, we used the quadratic formula because factoring was not straightforward.

The quadratic formula is a universal method to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our example, we identified the coefficients as follows:

• \( a = 2 \)
• \( b = -15 \)
• \( c = 7 \)

We substituted these values into the formula and simplified:\[ a = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 7}}{2 \cdot 2} \]
Simplify inside the square root:
\[ a = \frac{15 \pm \sqrt{225 - 56}}{4} \]
Simplify further to get the roots:\[ a = \frac{15 \pm \sqrt{169}}{4} = \frac{15 \pm 13}{4} \]
This results in two possible values for *a*:

• \( a = \frac{15 + 13}{4} = 7 \)
• \( a = \frac{15 - 13}{4} = \frac{1}{2} \)

Factoring a quadratic equation involves rewriting it as a product of two binomials. This method can be quicker, but it only works if the quadratic equation can be factored easily. For example, if we have an equation like:
\( x^2 - 5x + 6 = 0 \)
We can factor it into:\( (x - 2)(x - 3) = 0 \)
This gives us the solutions \( x = 2 \) and \( x = 3 \). Unfortunately, in our original problem, the equation \( 2a^2 - 15a + 7 = 0 \) does not factor easily into simple binomials. Hence, we used the quadratic formula instead.
However, it's always good to try factoring first because it can be simpler and faster. Checking for possible factorizations can sometimes save a lot of time. Always inspect whether factoring is feasible before moving on to more complex methods.