Quadratic equations and functions

$ax^2+bx+c=0$, where $a\ne 0$.

Sometimes, you have the quadratic equation without the $x$ term ($b=0$) or without the constant term ($c=0$).

Solving a quadratic equation

Solving a quadratic equation by completing the square.

Square root property

$x=\pm \sqrt{k}$

If $k$ is a positive number, then $x$ is a real number.

If $k$ is a negative number, then $x$ is an imaginary number.

Problem: 1 Solve $(x+9)^2-7=0$

Solution:

We have a square term with the variable, $(x+9)^2$ and a constant term, $-7$. First isolate the square term by adding $7$ on both sides,

$(x+9)^2=7$

Now use the square root property,

$x+9=\pm\sqrt{7}$

$x=-9\pm\sqrt{7}$

So, we have two values of $x$, one with the $+$ sign and the other with the $-$ sign:

ie, $x=-9+\sqrt{7}$; $x=-9-\sqrt{7}$

Problem 2: Solve the equation by completing the square.

   $x^2-6x-8=0$

Solution:

To complete the square, keep the variable terms on one side and move the constant term to the other side,

$x^2-6x=8$

Now square the half of the coefficient of $x$ and add it on both sides of the equation,

$x^2-6x+3^2=8+3^2$

Now, the left hand side is in the form, $a^2-2ab+b^2$ with $a=x$ and $b=3$. This we can write as $(a-b)^2$. Therefore,

$(x-3)^2=8+9$

$(x-3)^2=17$

Use the square root property,

$x-3=\pm \sqrt{17}$

$x=3\pm \sqrt{17}$

Solving a quadratic equation by quadratic formula

Deriving the quadratic formula

$ax^2+bx+c=0$

Move the constant term to the right hand side by adding $-c$ on both sides,

$ax^2+bx=-c$

Divide $a$, the coefficient of $x^2$ on both sides,

$x^2+\dfrac{b}{a}x=-\dfrac{c}{a}$

Now do completing the square, take the square of half of the coefficient of $x$ and add it on both sides,

$x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2=-\dfrac{c}{a}+\left(\dfrac{b}{2a}\right)^2$

Now the left hand side can be written as a perfect square,

$\left(x+\dfrac{b}{2a}\right)^2 =-\dfrac{c}{a}+\dfrac{b^2}{4a^2}$

  $=\dfrac{-4ac+b^2}{4a^2}$

Writing the positive term first in the numerator,

$\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}$

Now use the square root property,

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

$\hphantom{000000}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}$

Subtracting $\dfrac{b}{2a}$ on both sides,

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

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

This is the quadratic formula to find the roots of a quadratic equation.

Example 1: Use the quadratic formula to solve $2x^2+6x+4=0$

Solution:

$a=$coefficient of $x^2$$ =2$, $b=$ coefficient of $x$ $ =6$ and $c=$ constant term$ =4$

The quadratic formula is

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

Substituting the values of $a,b$ and $c$, we get

$x=\dfrac{-6\pm \sqrt{6^2-4\cdot 2\cdot 4}}{2\cdot 2}$

$x=\dfrac{-6\pm \sqrt{36-32}}{4}$

$x=\dfrac{-6\pm \sqrt{4}}{4}$

$x=\dfrac{-6\pm 2}{4}$

$x=\dfrac{-6\pm 2}{4}$

$x=\dfrac{-3\pm 1}{2}$

So, the solution is

$x=\dfrac{-3+1}{2}$ or $x=\dfrac{-3-1}{2}$

$x=-1$; $x=-2$

Example 2: Use the quadratic formula to solve $x^2=2x+4$

Solution:

$x^2-2x-4=0$

The quadratic formula is

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

$a=1$, $b=-2$ and $c=-4$. Substituting these values in the quadratic formula,

$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4\cdot 1\cdot (-4)}}{2\cdot 1}$

$\hphantom{x}=\dfrac{2\pm \sqrt{4+16}}{2}$

$\hphantom{x}=\dfrac{2\pm \sqrt{20}}{2}$

$\hphantom{x}=\dfrac{2\pm \sqrt{4\cdot 5}}{2}$

$\hphantom{x}=\dfrac{2\pm 2\sqrt{5}}{2}$

$\hphantom{x}=1\pm \sqrt{5}$

Example 3: Use the quadratic formula to solve $3x^2-5x=-7$

Solution:

$3x^2-5x+7=0$

The quadratic formula is

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

$a=3$, $b=-5$ and $c=7$. Substituting these values in the quadratic formula,

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4\cdot 3\cdot 7}}{2\cdot 3}$

$\hphantom{x}=\dfrac{5\pm \sqrt{25-84}}{6}$

$\hphantom{x}=\dfrac{5\pm i\sqrt{59}}{6}$

$\hphantom{x}=\dfrac{5}{6}\pm i\dfrac{\sqrt{59}}{6}$

Example 4: Use the quadratic formula to solve $x^4-15x^2+50=0$

Solution:

Take, $x^2=y$

So, $x^4=(x^2)^2=y^2$

Substituting $x^2$ and $x^4$ in the given equation,

$y^2-15y+50=0$

This is a quadratic equation, that we can solve for $y$ using the quadratic formula,

$y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=1,b=-15$ and $c=50$

Substituting these values,

$y=\dfrac{15\pm \sqrt{(-15)^2-4\cdot 1\cdot 50}}{2\cdot 1}$

$y=\dfrac{15\pm \sqrt{225-200}}{2}$

$y=\dfrac{15\pm \sqrt{25}}{2}$

$y=\dfrac{15\pm 5}{2}$

You have two solutions for $y$,

$y=\dfrac{15+ 5}{2}$; $y=\dfrac{15- 5}{2}$

$y=10$; $y=5$

Since we need the $x$ values, for each $y$ value we need to find the $x$ values.

For, $y=10$, substitute this in the $x^2=y$ equation,

$x^2=10$

Use the square root property,

$x=\pm \sqrt{10}$

i.e., $x=\sqrt{10}$; $x=-\sqrt{10}$

Next, take the other $y$ value and find the $x$,

$x^2=5$

$x=\pm \sqrt{5}$

$x= \sqrt{5}$; $x= -\sqrt{5}$

In total we have four $x$ values as the solution,

$x=\sqrt{10}$; $x=-\sqrt{10}$; $x= \sqrt{5}$; $x= -\sqrt{5}$

Applications of quadratic formula.

Example 1:A rock was thrown upward with an initial velocity of $42 ft/s$. The height, $h$ of the rock in feet after 𝑡 seconds is given by

$h=42t-16t^2$

Find the value(s) of 𝑡 for which the height of the rock is $27 ft$. Round your answer(s) to the nearest hundredth.

Solution:

The height is measured from the ground. The height of the rock increases as it goes up until it reaches the maximum height. Then it decreases and it becomes zero as the rock reaches the ground.

The question is asking to find the time when the rock is at a height of $27ft$ .

Take $h=27$ and solve the equation,

$27=42t-16t^2$

Let us move the terms on the right hand side to the left hand side by adding, $-42t+16t^2$ on both sides, so that we will have the quadratic equation in the standard form,

$16t^2-42t+27=0$

Now use the quadratic formula to solve the equation. Use the variable $t$ instead of $x$.

$t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=16, b=-42$ and $c=27$

Substituting these values in the quadratic formula,

$t=\dfrac{42\pm \sqrt{(-42)^2-4\cdot 16\cdot 27}}{2\cdot 16}$

$t=\dfrac{42\pm \sqrt{1764-1728}}{32}$

$t=\dfrac{42\pm \sqrt{36}}{32}$

$t=\dfrac{42\pm 6}{32}$

$t=\dfrac{42+6}{32}$ or $t=\dfrac{42-6}{32}$

$t=\dfrac{48}{32}$ or $t=\dfrac{36}{32}$

Usig a calculator and rounding the result to the nearest hundredth,

$t=1.50$ or $t=1.13$

We have two times, $t=1.50 s$ and $t= 1.13s$

The smaller time $1.13s$ is the time the rock reaches the height of $27 ft$ when it goes upward. And the other time $1.50s$ is time when the height of the ball is again at $27ft$, this happens during the return journey.

Example 2: A ball is thrown downward from a height of $221ft$ with an initial velocity of $12 ft/s$. The height of the ball after $t$ seconds is given by $h=221-12t-16t^2$. Determine how long it takes the ball to hit the ground?

Solution:

Here a ball is thrown downward from a height of $221 ft$. As the ball comes down, its height decreases and when it hits the ground the height becomes zero.

Since we need to find the time, as it hits the ground, you need to set $h=0$ in the equation and solve for $t$.

$0=221-12t-16t^2$

Let us make the $t^2$ term positive by moving everything to the left hand side,

$16t^2+12t-221=0$

$a=16$, $b=12$ and $c=-221$

Putting these values in the quadratic formula,

$t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t=\dfrac{-12\pm \sqrt{12^2-4\cdot 16\cdot (-221)}}{2\cdot 16}$

$t=\dfrac{-12\pm \sqrt{144+14144}}{32}$

$t=\dfrac{-12\pm \sqrt{14288}}{32}$

Use a calculator, $\sqrt{14288}=119.532$ (keep at least one more digit than the question demands). Substituting this value,

$t=\dfrac{-12\pm 119.532}{32}$

We have two values for $t$,

$t=\dfrac{-12+119.532}{32}$; $t=\dfrac{-12-119.532}{32}$

$t=\dfrac{-12+119.532}{32}$; $t=\dfrac{-12-119.532}{32}$

$t=3.36$; $t=-4.11$

We have a positive value and a negative for $t$. Since time cannot be negative, the answer is the positive time.

i.e., $t=3.36 s$

Discriminant

Discriminant $=b^2-4ac$

We can use the discriminant to determine the number of solutions, and the type of the solutions of a quadratic equation.

If the discriminant is not a perfect square, then the solutions are irrational.