Solving Quadratic Equations Using the Quadratic Formula

Solving Quadratic Equations Using the Quadratic Formula exercise

• Factor the quadratic equation.

  Hints

  A standard quadratic equation is of the form $ax^2+bx+c=0$.

  Pay attention to the signs of $a$, $b$, and $c$.

  To factor any quadratic equation of the form $x^2+bx+c$, find the factors of $c$ which sum to $b$.

  The zero factor property states that if any product equals zero then one of its factors must also equal zero.

  Solution

  To factor the quadratic equation $-1x^2-3x+4=0$, we must first factor out $-1$ to get

  $-1(x^2+3x-4)=-(x^2+3x-4)$.

  Next we're looking for the factors of $-4$ which sum to $3$:

  • $1\times (-4)=-4$ and $1-4=-3$ This pair doesn't work.
  • $-1\times 4=-4$ and $-1+4=3$ $~~~~~$✓

  Thus we can conclude that

  $-x^2-3x+4=-(x^2+3x-4)=-(x+4)(x-1)$.

  To solve the equation $-x^2-3x+4=0$ we can also solve $-(x+4)(x-1)=0$.

  Using the zero factor property, we get either

  $\begin{array}{rcl} x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x & = & ~-4 \end{array}$

  or

  $\begin{array}{rcl} x-1 & = & ~0\\ \color{#669900}{+1} & &\color{#669900}{+1}\\ x & = & ~1 \end{array}$

• Use the quadratic formula to solve the equation.

  Hints

  $ax^2+bx+c=0$ is the standard form of a quadratic equation.

  You still have to put the values of $a$, $b$, and $c$ into the quadratic formula.

  The first step is $x=\frac{-30\pm\sqrt{30^2-4(-1)(-100)}}{2(-1)}$.

  Solution

  In the equation above we have $a=-1$, $b=30$ and $c=-100$.

  We can use the quadratic formula to factor this equation. The quadratic formula is given by:

  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

  Plugging $a$, $b$, and $c$ into it gives us

  $x=\frac{-30\pm\sqrt{30^2-4(-1)(-100)}}{2(-1)}$.

  Next we simplify to get

  $\begin{array}{rcl} x&=&~\frac{-30\pm\sqrt{30^2-4(-1)(-100)}}{2(-1)}\\ &=&~-\frac{-30\pm\sqrt{900-400}}{2}\\ &=&~-\frac{-30\pm\sqrt{500}}{2}\\ x_1&=&~-\frac{-30+\sqrt{500}}{2}\approx 3.82\\ x_2&=&~-\frac{-30-\sqrt{500}}{2}\approx26.18\\ \end{array}$

• Find the value of $x$ which satisfies each quadratic equation.

  Hints

  First identify $a$, $b$, and $c$.

  $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the coefficient of $1$.

  Pay attention to signs.

  Solution

  Once you have identified $a$, $b$, and $c$ in the quadratic equation $ax^2+bax+c=0$, you can put these values into the quadratic formula:

  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

  $~$

  For $x^2+5x+4=0$ $\rightarrow$ $a=1$, $b=5$ and $c=4$

  The solutions are given by

  $x=\frac{-5\pm\sqrt{5^2-4(1)(4)}}{2(1)}$.

  $~$

  For $-x^2-4x+5=0$ $\rightarrow$ $a=-1$, $b=-4$ and $c=5$

  The solutions are given by

  $x=\frac{4\pm\sqrt{(-4)^2-4(-1)(5)}}{2(-1)}$.

  $~$

  For $2x^2-4x-16=0$ $\rightarrow$ $a=2$, $b=-4$ and $c=-16$

  The solutions are given by

  $x=\frac{4\pm\sqrt{(-4)^2-4(2)(-16)}}{2(-1)}$.

  $~$

  For $2x^2+8x-10=0$ $\rightarrow$ $a=2$, $b=8$ and $c=-10$

  The solutions are given by

  $x=\frac{-8\pm\sqrt{8^2-4(2)(-10)}}{2(2)}$.

• Determine the path of the quadratic equation.

  Hints

  For $x^2+2x+1=0$, we have that $a=1$, $b=2$, and $c=1$.

  Use the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

  One solution is negative and the other one is positive.

  Solution

  We have to solve the equation $-0.2x^2+x+10=0$ with $a=-0.2$, $b=1$ and $c=10$.

  We put these values for $a$, $b$, and $c$ into the quadratic formula,

  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$,

  to get

  $x=\frac{-1\pm\sqrt{1^2-4(-0.2)(10)}}{2(-0.2)}$.

  Then we simplify:

  $\begin{array}{rcl} x&=&~\frac{-1\pm\sqrt{1^2-4(-0.2)(10)}}{2(-0.2)}\\ &=&~\frac{-1\pm\sqrt{1+8}}{-0.4}\\ &=&~-\frac{-1\pm\sqrt9}{0.4}\\ x_1&=&~-\frac{-1+3}{0.4}=-5\\ &=&~-\frac{-1-3}{0.4}=10 \end{array}$

  We can graph these solutions as the $x$-intercepts of the graph of the quadratic equation, which will look like a downwards opened parabola (as $a<0$).

• Identify the values of $a$, $b$, and $c$ for each quadratic equation.

  Hints

  For any quadratic equation $ax^2+bx+c=0$, keep the following in mind:

  • $a$ is the coefficient of $x^2$
  • $b$ is the coefficient of $x$
  • $c$ is the coefficient of $1$

  Pay attention to signs.

  For $-3x^2+x+5=0$, we have:

  • $a=-3$
  • $b=1$
  • $c=5$

  Solution

  Let's use the quadratic formula:

  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

  To solve a quadratic equation $ax^2+bx+c=0$ using the quadratic formula, we must identify $a$, $b$, and $c$ in the corresponding equation:

  • $a$ is the coefficient of $x^2$
  • $b$ is the coefficient of $x$
  • $c$ is the coefficient of $1$

  For each of these values, you have to keep the sign in mind.

  Let's identify $a$, $b$, and $c$ for two examples:

  $-1x^2+(-3)x+4=0$

  • $a=-1$
  • $b=-3$
  • $c=4$

  $-1x^2+30x+(-100)=0$

  • $a=-1$
  • $b=30$
  • $c=-100$

• Solve the given quadratic equations.

  Hints

  You find the solution of the first equation by simplifying $x=\frac{-5\pm\sqrt{5^2-4(1)(4)}}{2(1)}$.

  You can also check the solutions by putting them into the original equation.

  Keep track of the signs of the coefficients.

  Solution

  First determine $a$, $b$, and $c$ in the quadratic equation $ax^2+bx+c=0$.

  Next put these values for $a$, $b$, and $c$ into the quadratic formula:

  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

  $~$

  Let's take a look at the quadratic equations we were asked to solve:

  $x^2+5x+4=0$ with $a=1$, $b=5$ and $c=4$.

  The solutions are given by

  $\begin{array}{rcl} x&=&~\frac{-5\pm\sqrt{5^2-4(1)(4)}}{2(1)}\\ &=&~\frac{-5\pm\sqrt{25-16}}{2}\\ &=&~\frac{-5\pm\sqrt9}{2}\\ x_1&=&~\frac{-5+3}{2}=-1\\ x_2&=&~\frac{-5-3}{2}=-4 \end{array}$

  $~$

  $-x^2-4x+5=0$ with $a=-1$, $b=-4$ and $c=5$.

  The solutions are given by

  $\begin{array}{rcl} x&=&~\frac{4\pm\sqrt{(-4)^2-4(-1)(5)}}{2(-1)}\\ &=&~\frac{4\pm\sqrt{16+20}}{-2}\\ &=&~-\frac{4\pm\sqrt{36}}{2}\\ x_1&=&~-\frac{4+6}{2}=-5\\ x_2&=&~-\frac{4-6}{2}=1 \end{array}$

  $~$

  $2x^2-4x-16=0$ with $a=2$, $b=-4$ and $c=-16$.

  The solutions are given by

  $\begin{array}{rcl} x&=&~\frac{4\pm\sqrt{(-4)^2-4(2)(-16)}}{2(2)}\\ &=&~\frac{4\pm\sqrt{16+128}}{4}\\ &=&~\frac{4\pm\sqrt{144}}{4}\\ x_1&=&~\frac{4+12}{4}=4\\ &=&~\frac{4-12}{4}=-2 \end{array}$

  $~$

  $2x^2+8x+6=0$ $\rightarrow$ $a=2$, $b=8$ and $c=6$

  The solutions are given by

  $\begin{array}{rcl} x&=&~\frac{-8\pm\sqrt{8^2-4(2)(6)}}{2(2)}\\ &=&~\frac{-8\pm\sqrt{64-48}}{4}\\ &=&~\frac{-8\pm\sqrt{16}}{4}\\ x_1&=&~\frac{-8+4}{4}=-1\\ x_2&=&~\frac{-8-4}{4}=-3 \end{array}$