Solve the equation:
$x^2+2x-35=0$
The given equation is a quadratic equation of the form \(ax^2 + bx + c = 0\), specifically \(x^2 + 2x - 35 = 0\). Here, the coefficients are \(a = 1\), \(b = 2\), and \(c = -35\). Our goal is to find the values of \(x\) that satisfy this equation. We can approach this by either factoring the quadratic trinomial or by using the quadratic formula.
We seek two numbers \(p\) and \(q\) such that their product \(pq = -35\) and their sum \(p + q = 2\). Looking at the factors of \(-35\), we consider pairs like \((-1, 35)\), \((1, -35)\), \((-5, 7)\), and \((5, -7)\). The pair \(-5\) and \(7\) satisfies the condition since \((-5) \times 7 = -35\) and \((-5) + 7 = 2\). Thus, the quadratic expression \(x^2 + 2x - 35\) can be factored as \((x - 5)(x + 7) = 0\).
According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero:
1) \(x - 5 = 0 \implies x = 5\)
2) \(x + 7 = 0 \implies x = -7\)
These are the two solutions to the equation.
To verify our results using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we substitute \(a=1\), \(b=2\), and \(c=-35\):
\[x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-35)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 140}}{2} = \frac{-2 \pm \sqrt{144}}{2} = \frac{-2 \pm 12}{2}\]
Calculating the two possible values:
1) \(x = \frac{-2 + 12}{2} = \frac{10}{2} = 5\)
2) \(x = \frac{-2 - 12}{2} = \frac{-14}{2} = -7\)
Both methods yield the same results, confirming that the solutions are \(x = 5\) and \(x = -7\). The final answer is \(\boxed{5, -7}\).
Quadratic equations are widely used in physics and engineering. For example, if you throw a ball into the air, its height \(h\) as a function of time \(t\) can often be described by a quadratic equation such as \(h(t) = -5t^2 + 10t + 1\). Finding when the ball hits the ground involves solving for \(t\) when \(h(t) = 0\), which is precisely the type of calculation performed here to find the roots of a quadratic equation. Mastery of this algebraic technique allows one to predict outcomes in motion, economics, and architectural design.