Solving $(x-7)(x+9)=0$: Which Equation Matches?

Are you ready to dive into the fascinating world of algebra and conquer quadratic equations? Today, we're tackling a common problem: finding an equation that shares the exact same solutions as a given factored form. Our starting point is the equation $(x-7)(x+9)=0$. This equation is already beautifully factored, which makes finding its solutions a breeze! The Zero Product Property tells us that if the product of two or more factors is zero, then at least one of the factors must be zero. So, for $(x-7)(x+9)=0$ to be true, either $(x-7)=0$ or $(x+9)=0$. Solving these simple linear equations, we get $x=7$ and $x=-9$. These are the two roots, or solutions, of our original equation. Now, the challenge is to identify which of the provided options, when expanded into standard quadratic form ($ax^2+bx+c=0$), will yield these very same solutions, $x=7$ and $x=-9$. This means we need to work backward from the solutions to reconstruct the quadratic equation, or alternatively, expand each option and find its roots to see which one matches.

Let's explore the first approach: reconstructing the equation from its roots. If we know the roots of a quadratic equation are $r_1$ and $r_2$, we can write the equation in factored form as $a(x-r_1)(x-r_2)=0$, where 'a' is any non-zero constant. Since our original equation has a leading coefficient of 1 (implied by the absence of any multiplier in front of the parentheses), we'll aim for an 'a' value of 1. Our roots are $7$ and $-9$. So, we can set up the factored form as $(x-7)(x-(-9))=0$, which simplifies to $(x-7)(x+9)=0$. This brings us right back to where we started! The next step, and the one that will help us compare with the given options, is to expand this factored form into the standard quadratic equation $ax^2+bx+c=0$. We can do this using the distributive property (often remembered by the acronym FOIL: First, Outer, Inner, Last).

Multiplying the terms in $(x-7)(x+9)$:

• First: $x * x = x^2$
• Outer: $x * 9 = 9x$
• Inner: $-7 * x = -7x$
• Last: $-7 * 9 = -63$

Combining these terms: $x^2 + 9x - 7x - 63$. Now, we simplify by combining the like terms (the 'x' terms): $9x - 7x = 2x$. This gives us the expanded quadratic equation: $x^2 + 2x - 63 = 0$.

So, the equation $(x-7)(x+9)=0$ is exactly equivalent to $x^2 + 2x - 63 = 0$. This means that option A is our correct answer! But to be thorough and to reinforce our understanding, let's quickly examine why the other options are incorrect. This process not only confirms our answer but also sharpens our algebraic skills.

Let's systematically analyze each option to confirm our finding and solidify our understanding of how quadratic equations relate to their solutions. We've already determined that the equation $(x-7)(x+9)=0$ expands to $x^2 + 2x - 63 = 0$. This matches option A perfectly. However, in a test scenario, you might be asked to verify or you might want to double-check your work. The most direct way to do this is to expand each of the other options and see if they yield the same equation or, alternatively, attempt to solve each option to see if they produce the roots $x=7$ and $x=-9$. Let's take the expansion route for options B, C, and D.

• Option B: $x^2+2 x+63=0$ This equation has the same $x^2$ and $x$ terms as our target equation ($x^2 + 2x - 63 = 0$), but the constant term is $+63$ instead of $-63$. This difference in the constant term is crucial. If we were to try and factor this, we would be looking for two numbers that multiply to $+63$ and add to $+2$. The pairs of factors for 63 are (1, 63), (3, 21), and (7, 9). None of these pairs, when added or subtracted, will result in 2. This indicates that this equation does not have simple integer roots like 7 and -9, and it certainly won't have the same solutions. The discriminant ($\Delta = b^2 - 4ac$) for this equation is $2^2 - 4(1)(63) = 4 - 252 = -248$. Since the discriminant is negative, this equation has two complex conjugate roots, not the real roots we found earlier.

• Option C: $x^2-2 x+63=0$ Here, the coefficient of the $x$ term has changed sign ($ -2x $ instead of $ +2x $), and the constant term is also $+63$. This sign change for the $x$ term implies that the sum of the roots would be positive (according to Vieta's formulas, the sum of roots is $-b/a$, so here it would be $-(-2)/1 = 2$). However, the constant term $+63$ is still problematic. If we tried to factor this, we'd need two numbers that multiply to $+63$ and add to $-2$. Again, looking at the factors of 63 (1, 63; 3, 21; 7, 9), no combination adds up to $-2$. The discriminant here is $(-2)^2 - 4(1)(63) = 4 - 252 = -248$, which again points to complex roots.

• Option D: $x^2-2 x-63=0$ This option has the correct constant term ($-63$), but the $x$ term is $-2x$ instead of $+2x$. If we were to factor this, we'd look for two numbers that multiply to $-63$ and add to $-2$. Let's consider the factors of $-63$: (1, -63), (-1, 63), (3, -21), (-3, 21), (7, -9), (-7, 9). Now let's check the sums: $1+(-63)=-62$, $-1+63=62$, $3+(-21)=-18$, $-3+21=18$, $7+(-9)=-2$, $-7+9=2$. Aha! The pair $(7, -9)$ adds up to $-2$. So, this equation factors into $(x+7)(x-9)=0$. The solutions to this equation would be $x=-7$ and $x=9$. These are not the same solutions as our original equation ($x=7$ and $x=-9$). This confirms that option D is also incorrect.

Let's summarize our findings. We started with $(x-7)(x+9)=0$, whose solutions are $x=7$ and $x=-9$. By expanding this expression, we found it is equivalent to $x^2 + 2x - 63 = 0$. This perfectly matches option A. We then analyzed options B, C, and D by considering their constant terms and coefficients, and even calculating their discriminants or attempting factorization. Each of these other options leads to different solutions or complex roots, confirming that they are not equivalent to the original equation. Therefore, the only equation that has exactly the same solutions as $(x-7)(x+9)=0$ is $x^2 + 2x - 63 = 0$. This detailed step-by-step process, involving expansion and verification, ensures a thorough understanding of why option A is the correct choice and reinforces key algebraic principles.

Understanding how to manipulate quadratic equations, whether by factoring, expanding, or using the quadratic formula, is a fundamental skill in algebra. The ability to recognize equivalent forms of an equation is just as important. In this case, we saw that the factored form $(x-7)(x+9)=0$ and the standard form $x^2+2x-63=0$ represent the same mathematical relationship and thus share the same solution set. This equivalence arises directly from the properties of polynomial multiplication and the Zero Product Property. When faced with similar problems, remember to either expand the factored form to match the standard form options or, conversely, factor the standard form options to see if they yield the original factors. Always pay close attention to the signs of the coefficients, as a single sign error can lead to entirely different solutions. Practicing these transformations will build your confidence and proficiency in algebra.

For further exploration into the world of quadratic equations and algebraic manipulation, you can refer to resources like Khan Academy's Algebra section, which offers comprehensive lessons and practice exercises on these topics. Additionally, Math is Fun's algebra page provides clear explanations and examples for a wide range of mathematical concepts.