### Home > CCAA8 > Chapter 9 Unit 10 > Lesson CCA: 9.1.4 > Problem9-39

9-39.

Solve the following quadratic equations using any method.

1. $10000x^2−64=0$

1. $9x^2−8=−34x$

1. $2x^2−4x+7=0$

1. $3.2x+0.2x^2−5=0$

Example 1: Solve $3x^2+x−14=0$ for $x$ using the Zero Product Property.

Solution: First, factor the quadratic so it is written as a product: $(3x+7)(x−2)=0$. (If factoring is not possible, one of the other methods of solving must be used.)  The Zero Product Property states that if the product of two terms is $0$, then at least one of the factors must be $0$.  Thus, $3x+7=0$ or $x−2=0$. Solving these equations for $x$ reveals that $x=\frac{7}{3}$ or that $x=2$.

Example 2: Solve $3x^2+x−14=0$ for $x$ using the Quadratic Formula.

Solution: This method works for any quadratic.  First, identify $a,b$, and $c$$a$ equals the number of $x^2$-terms, $b$ equals the number of $x$-terms, and $c$ equals the constant. For $3x^2+x−14=0,a=3,b=1$, and $c=−14$.  Substitute the values of $a,b$, and $c$ into the Quadratic Formula and evaluate the expression twice: once with addition and once with subtraction. Examine this method below:

 $x=\frac{-1+\sqrt{1^2-4(3)(-14)}} {2⋅3}$ $x=\frac{-1+\sqrt{1^2-4(3)(-14)}} {2⋅3}$ $=\frac{-1+\sqrt{169}} {6}$ or $=\frac{-1-\sqrt{169}}{6}$ $=\frac{12}{6}=2$ $=\frac{-14}{6}=-\frac{7}{3}$

Example 3: Solve $x^2+5x+4=0$ by completing the square.

Solution: This method works most efficiently when the coefficient of $x^2$ is $1$.  Rewrite the equation as $x^2+5x=−4$.  Rewrite the left side as an incomplete square:

 Generic rectangle left edge$2.5$+$x$ interior top left$2.5x$ blank interior bottom left$x^2$ interior bottom right$2.5x$ right edge$=-4$ bottom edge $x\ +\ 2.5$

Complete the square and rewrite as
$(x+2.5)^2−6.25=−4$ or

$(x+2.5)^2=2.25$

Take the square root of both sides, $x+2.5=\pm1.5$.  Solving for $x$ reveals that $x=−1$ or $x=−4$.