Quadratic Equations & Applications: Videos & Practice Problems

The quadratic equation, expressed in $a{x}^2+bx+c=0$ standard form, models various real-world scenarios. Solving involves factoring the trinomial, setting each binomial factor equal to zero, and applying the zero product property. This process identifies values of $x$ that satisfy the equation. Understanding coefficients, terms, and factoring techniques is essential for mastering polynomial equations and their applications in algebra and beyond.

Solving Quadratic Equations by Factoring

The quadratic equation is a fundamental mathematical tool used to model various real-world scenarios, such as the trajectory of projectiles and analyzing business sales versus costs. It is expressed in the standard form as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable to solve for. Solving a quadratic equation involves finding the values of \(x\) that satisfy this equation, making it equal to zero.

To solve a quadratic equation, the first step is to ensure it is written in standard form. Next, the equation is factored into two binomials. Factoring involves identifying two numbers that multiply to the constant term \(c\) and add up to the coefficient \(b\). For example, in the quadratic \(x^2 + 10x + 21 = 0\), the numbers 3 and 7 multiply to 21 and add to 10, allowing the equation to be factored as \((x + 3)(x + 7) = 0\).

Once factored, the zero product property is applied. This property states that if the product of two factors equals zero, then at least one of the factors must be zero. Therefore, setting each factor equal to zero gives the equations \(x + 3 = 0\) and \(x + 7 = 0\). Solving these yields the solutions \(x = -3\) and \(x = -7\).

Verification of solutions is essential. Substituting \(x = -3\) back into the original equation results in \((-3)^2 + 10(-3) + 21 = 9 - 30 + 21 = 0\), confirming it as a valid solution. Similarly, substituting \(x = -7\) gives \((-7)^2 + 10(-7) + 21 = 49 - 70 + 21 = 0\), also confirming its validity.

Understanding how to factor and solve quadratic equations is crucial for tackling a wide range of mathematical problems. Mastery of this process enhances problem-solving skills and prepares learners for more advanced algebraic concepts.

Solving Quadratic Equations by Factoring Example 1

To solve a quadratic equation by factoring, the first step is to write the equation in standard form, which means having all terms on one side set equal to zero. For example, if the equation is not already in standard form, you can rearrange it by adding or subtracting terms on both sides. Once the quadratic is in the form \(ax^2 + bx + c = 0\), you can proceed to factor it.

When the leading coefficient \(a\) is not equal to 1, the ac method is an effective factoring strategy. This involves identifying the coefficients \(a\), \(b\), and \(c\) from the quadratic. Then, multiply \(a\) and \(c\) to find the product \(ac\). Next, find two numbers that multiply to \(ac\) and add up to \(b\). These two numbers will help break apart the middle term.

For instance, if \(a = -1\), \(b = -3\), and \(c = 4\), then \(ac = -4\). The factor pairs of \(-4\) are \((1, -4)\) and \((-1, 4)\). Since the sum must be \(-3\), the correct pair is \((1, -4)\). Using these, rewrite the quadratic by splitting the middle term: \(-x^2 + x - 4x + 4 = 0\).

Next, apply factoring by grouping. Group the terms into two pairs: \((-x^2 + x)\) and \((-4x + 4)\). Factor out the greatest common factor (GCF) from each group. From the first group, factor out \(x\), leaving \(-x + 1\). From the second group, factor out \(-4\), also leaving \(-x + 1\). Since both groups contain the common binomial factor \((-x + 1)\), factor this out to get:

\[(-x + 1)(x - 4) = 0\]

Finally, solve for \(x\) by setting each factor equal to zero:

\[-x + 1 = 0 \implies x = 1\]\[x - 4 = 0 \implies x = 4\]

These values, \(x = 1\) and \(x = 4\), are the solutions to the quadratic equation. To verify, substitute these values back into the original equation to ensure they satisfy it. This method highlights the importance of rewriting equations in standard form, using the ac method for factoring when \(a \neq 1\), and applying factoring by grouping to efficiently solve quadratic equations.

Use the zero product rule to solve the following equations for $x$ .
$(x-5)(x+3)=0$

• A. $x=-5$ or $x=-3$
• B. $x=5$ or $x=3$
• C. $x=-5$ or $x=3$
• D. $x=5$ or $x=-3$

Answer: (D)

Use the zero product rule to solve the following equations for $x$ .
$2(x-4)(x+6)=0$

• A. $x=4$ or $x=-6$
• B. $x=-4$ or $x=6$
• C. $x=4$ or $x=6$
• D. $x=-4$ or $x=-6$

Answer: (A)

Solve the following quadratic equations.
$2x^2-10x=0$

• A. $x=0$ or $x=-5$
• B. $x=2$ or $x=-5$
• C. $x=0$ or $x=5$
• D. $x=0$ or $x=10$

Answer: (C)

Solve the following quadratic equations.
$2x^2=5x+3$

• A. $x=\frac12$ or $x=-3$
• B. $x=-\frac12$ or $x=3$
• C. $x=-\frac12$ or $x=-3$
• D. $x=\frac12$ or $x=3$

Answer: (B)

Here’s what students ask on this topic:

What is the standard form of a quadratic equation and why is it important?

The standard form of a quadratic equation is $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants and $a\backslash neq0$. This form is important because it provides a clear structure to solve the equation by factoring, completing the square, or using the quadratic formula. Writing the equation in standard form allows us to identify the coefficients and constant term, which are essential for applying these solution methods. It also helps in modeling real-world problems, such as projectile motion or business profit analysis, by representing relationships in a consistent way.

How do you factor a quadratic equation to solve for x?

To factor a quadratic equation like $x{2}^{}+10x+21=0$, first find two numbers that multiply to the constant term ($21$) and add to the coefficient of $x$ ($10$). Here, $3$ and $7$ work because $3\backslash times7=21$ and $3+7=10$. Then, rewrite the quadratic as $(x+3)(x+7)\; =0$. Using the zero product property, set each factor equal to zero: $x+3=0$ or $x+7=0$. Solving these gives $x=\; -3$ or $x=\; -7$, which are the solutions.

What is the zero product property and how is it used in solving quadratic equations?

The zero product property states that if the product of two factors is zero, then at least one of the factors must be zero. Mathematically, if $A\backslash timesB=0$, then either $A=0$ or $B=0$. This property is crucial when solving quadratic equations after factoring. Once the quadratic is factored into two binomials, such as $(x+3)(x+7)\; =0$, we apply the zero product property to set each factor equal to zero. This gives two simple linear equations to solve for $x$. It simplifies finding the roots of the quadratic equation efficiently.

How can you check if the solutions to a quadratic equation are correct?

To check solutions of a quadratic equation, substitute each solution back into the original equation and verify if the equation equals zero. For example, if the solutions are $x=\; -3$ and $x=\; -7$ for the equation $x{2}^{}+10x+21=0$, plug in $-3$: $(-3)^2\; +\; 10(-3)\; +\; 21\; =\; 9\; -\; 30\; +\; 21\; =\; 0$. Then plug in $-7$: $(-7)^2\; +\; 10(-7)\; +\; 21\; =\; 49\; -\; 70\; +\; 21\; =\; 0$. Since both equal zero, the solutions are correct. This step confirms the accuracy of your factoring and solving process.

Why is factoring preferred when solving some quadratic equations?

Factoring is often preferred for solving quadratic equations because it is a straightforward and efficient method when the quadratic trinomial can be easily factored into binomials. It allows you to quickly find the roots by applying the zero product property without needing more complex methods like completing the square or the quadratic formula. Factoring also helps reinforce understanding of polynomial structure, coefficients, and the relationship between factors and roots. However, it works best when the quadratic factors nicely with integer roots, making it a practical first approach in many cases.