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

x = 5 x=5 x = 5 or x = − 3 x=-3 x = − 3

x = − 5 x=-5 x = − 5 or x = − 3 x=-3 x = − 3

x = 5 x=5 x = 5 or x = 3 x=3 x = 3

x = − 5 x=-5 x = − 5 or x = 3 x=3 x = 3

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

x = 4 x=4 x = 4 or x = − 6 x=-6 x = − 6

x = − 4 x=-4 x = − 4 or x = 6 x=6 x = 6

x = 4 x=4 x = 4 or x = 6 x=6 x = 6

x = − 4 x=-4 x = − 4 or x = − 6 x=-6 x = − 6

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

x = 0 x=0 x = 0 or x = 5 x=5 x = 5

x = 0 x=0 x = 0 or x = − 5 x=-5 x = − 5

x = 2 x=2 x = 2 or x = − 5 x=-5 x = − 5

x = 0 x=0 x = 0 or x = 10 x=10 x = 10

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

x = − 1 2 x=-\frac12 x = − 2 1 or x = 3 x=3 x = 3

x = 1 2 x=\frac12 x = 2 1 or x = − 3 x=-3 x = − 3

x = − 1 2 x=-\frac12 x = − 2 1 or x = − 3 x=-3 x = − 3

x = 1 2 x=\frac12 x = 2 1 or x = 3 x=3 x = 3

7. Factoring

Quadratic Equations & Applications

7. Factoring

Quadratic Equations & Applications: Videos & Practice Problems

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Topic summary

The quadratic equation, expressed in $a x^{2} + b x + c = 0$ standard form, models various real-world scenarios. Solving involves factoring the trinomial into binomials, then applying the zero product property to find values of $x$ that satisfy the equation. This process highlights key polynomial concepts like coefficients, terms, and exponents. Factoring requires identifying numbers that multiply to $c$ and add to $b$ . Verifying solutions ensures accuracy, reinforcing understanding of polynomial structure and solution methods.

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Solving Quadratic Equations by Factoring

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Solving Quadratic Equations by Factoring Video Summary

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.

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Solving Quadratic Equations by Factoring Example 1

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Solving Quadratic Equations by Factoring Example 1 Video Summary

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.

Problem

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

$x=-5$ or $x=-3$

$x=5$ or $x=3$

$x=-5$ or $x=3$

$x=5$ or $x=-3$

Problem

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

$x=4$ or $x=-6$

$x=-4$ or $x=6$

$x=4$ or $x=6$

$x=-4$ or $x=-6$

Problem

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

$x=0$ or $x=-5$

$x=2$ or $x=-5$

$x=0$ or $x=5$

$x=0$ or $x=10$

Problem

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

$x=\frac12$ or $x=-3$

$x=-\frac12$ or $x=3$

$x=-\frac12$ or $x=-3$

$x=\frac12$ or $x=3$

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Applications of Quadratic Equations

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Applications of Quadratic Equations Example 2

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Applications of Quadratic Equations Example 3

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Applications of Quadratic Equations Example 4

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Applications of Quadratic Equations Example 5

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Applications of Quadratic Equations Example 6

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The standard form of a quadratic equation is $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants and $\neq 0$ . This form is important because it organizes the equation by descending powers of $x$ , making it easier to apply methods like factoring, completing the square, or using the quadratic formula to find solutions. Writing the equation in standard form ensures consistency and clarity, which is essential for solving and analyzing quadratic equations in various real-world applications such as projectile motion or business cost analysis.

To factor a quadratic equation like $a x^{2} + b x + c = 0$ , first ensure it is in standard form. Then, find two numbers that multiply to $c$ and add to $b$ . These numbers help break the middle term and factor the trinomial into two binomials, for example, $(x + m)(x + n) = 0$ . Next, apply the zero product property, setting each factor equal to zero: $x + m = 0$ and $x + n = 0$ . Solving these gives the values of $x$ that satisfy the equation. This method is efficient for quadratics that factor neatly and is foundational for understanding polynomial behavior.

The zero product property states that if the product of two factors equals zero, then at least one of the factors must be zero. Mathematically, if $AB = 0$ , then either $A = 0$ or $B = 0$ . In solving quadratic equations, after factoring the quadratic into binomials, such as $(x + m)(x + n) = 0$ , we use this property to set each factor equal to zero: $x + m = 0$ and $x + n = 0$ . Solving these linear equations gives the roots of the quadratic. This property is crucial because it transforms a quadratic equation into simpler linear equations that are easier to solve.

To verify solutions of a quadratic equation, substitute each solution back into the original equation. For example, if the quadratic is $x^2 + 10x + 21 = 0$ and the solutions are $x = -3$ and $x = -7$ , plug these values into the equation. For $x = -3$ , calculate $(-3)^2 + 10(-3) + 21$ . This equals $9 - 30 + 21 = 0$ , confirming it is a solution. Repeat for $x = -7$ . If the equation equals zero for the substituted value, the solution is correct. This step ensures accuracy and reinforces understanding of the relationship between roots and the quadratic equation.

To solve a quadratic equation by factoring, follow these steps: (1) Write the equation in standard form $ax^2 + bx + c = 0$ . (2) Factor the quadratic trinomial into two binomials, finding numbers that multiply to $c$ and add to $b$ . (3) Apply the zero product property by setting each factor equal to zero. (4) Solve each resulting linear equation for $x$ . (5) Check your solutions by substituting them back into the original equation. This method is effective when the quadratic factors easily and provides a clear path to finding the roots.