Show how to find the zero(s): y=2x2+7x

List smaller solution first.
and

1 answer

To find the zeros of the quadratic function \( y = 2x^2 + 7x \), we need to set the equation equal to zero and solve for \( x \):

\[ 2x^2 + 7x = 0 \]

First, we can factor out the common term \( x \):

\[ x(2x + 7) = 0 \]

This gives us two factors to solve:

  1. \( x = 0 \)
  2. \( 2x + 7 = 0 \)

Now, let's solve the second equation:

\[ 2x + 7 = 0 \implies 2x = -7 \implies x = -\frac{7}{2} \]

Thus, the two solutions (or zeros) of the equation are:

  1. \( x = -\frac{7}{2} \)
  2. \( x = 0 \)

Listing the smaller solution first, the zeros of the function \( y = 2x^2 + 7x \) are:

\[ -\frac{7}{2}, 0 \]