- Rewrite the equation \(2v^2 = -5v - 3\) as \(2v^2 + 5v + 3 = 0\).
- Factor the quadratic expression into \((2v+3)(v+1) = 0\).
- Set each factor to zero to obtain the solutions \(v = -1.5\) and \(v = -1\).
- Final answer: \(\boxed{-1.5, -1}\).
Explanation
- Rewrite in Standard Form
To solve the quadratic equation \(2v^2 = -5v - 3\), we first rewrite it in the standard form \(av^2 + bv + c = 0\). By adding \(5v\) and \(3\) to both sides of the equation, we obtain:
\[2v^2 + 5v + 3 = 0\]
Here, the coefficients are \(a = 2\), \(b = 5\), and \(c = 3\).
- Solve by Factoring
We can solve this quadratic equation by factoring. We look for two numbers that multiply to \(a \times c = 2 \times 3 = 6\) and add up to \(b = 5\). These two numbers are 2 and 3. We can split the middle term \(5v\) into \(2v + 3v\):
\[2v^2 + 2v + 3v + 3 = 0\]
Now, group the terms and factor them:
\[(2v^2 + 2v) + (3v + 3) = 0\]
\[2v(v + 1) + 3(v + 1) = 0\]
Factor out the common binomial \((v + 1)\):
\[(2v + 3)(v + 1) = 0\]
- Finding the Solutions
Setting each factor equal to zero, we find the roots:
For \(2v + 3 = 0\), we subtract 3 from both sides: \(2v = -3\), then divide by 2: \(v = -\frac{3}{2} = -1.5\).
For \(v + 1 = 0\), we subtract 1 from both sides: \(v = -1\).
Thus, the solutions are \(v = -1.5, -1\). The final answer is \(\boxed{-1.5, -1}\).
Examples
Quadratic equations appear frequently in physics, for example, when calculating the time it takes for an object thrown into the air to return to the ground. If the height \(h\) at time \(t\) is given by \(h(t) = -16t^2 + v_0t + h_0\), setting \(h(t) = 0\) allows us to find the specific times the object hits the ground. Solving such equations is a fundamental skill in motion analysis and engineering.