Verified Answer :

  • Factor the expression \(x^2 + 3x - 4 = 0\) as \((x+4)(x-1) = 0\).
  • Apply the zero-product property to find the roots \(x = -4\) and \(x = 1\).
  • Verify the roots using the quadratic formula with \(a=1, b=3, c=-4\).
  • Identify the correct options from the list as \(-4\) and \(1\), so the final answer is \(\boxed{{-4, 1}}\).

Explanation

  1. Analysis of the Quadratic Equation

We are presented with the quadratic equation \(x^2 + 3x - 4 = 0\). Our objective is to identify the solutions to this equation from a provided list of numbers. A quadratic equation of the form \(ax^2 + bx + c = 0\) can generally be solved either by factoring the expression into two linear binomials or by using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). We will use both methods to ensure the accuracy of our results.

  1. Solving via Factoring

To factor the quadratic expression \(x^2 + 3x - 4\), we look for two numbers that multiply to the constant term \(-4\) and add to the coefficient of the linear term \(3\). Testing pairs of factors for \(-4\):

  • \(4\) and \(-1\): \(4 \times (-1) = -4\) and \(4 + (-1) = 3\).

Since these numbers satisfy our conditions, we can write the quadratic equation as:
\[(x + 4)(x - 1) = 0\]
By the zero-product property, the expression equals zero when either factor is zero:

  • \(x + 4 = 0 \implies x = -4\)
  • \(x - 1 = 0 \implies x = 1\)

Thus, the roots of the equation are \(-4\) and \(1\).

  1. Verification and Final Conclusion

We can also verify our solutions using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 3, c = -4\):

  • First, calculate the discriminant: \(D = b^2 - 4ac = (3)^2 - 4(1)(-4) = 9 + 16 = 25\).
  • Now, substitute into the formula: \(x = \frac{-3 \pm \sqrt{25}}{2(1)} = \frac{-3 \pm 5}{2}\).
  • Solving for both cases:
  • \(x_1 = \frac{-3 + 5}{2} = \frac{2}{2} = 1\)
  • \(x_2 = \frac{-3 - 5}{2} = \frac{-8}{2} = -4\)

The results are consistent with the factoring method. Comparing these values with the options provided (\(-4, 2, 4, 0, -1, -2, 1\)), we see that \(-4\) and \(1\) are indeed the correct choices.

Examples

Quadratic equations are widely used in physics and engineering to model motion. For instance, if you throw a ball, its height \(h\) over time \(t\) can be described by a quadratic equation like \(h(t) = -5t^2 + 20t + 1\). By setting \(h(t) = 0\), you can solve for the time at which the ball hits the ground using the same techniques used to solve our equation. Mastering these methods allows you to determine critical points in various real-world scenarios.

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