Solving a quadratic equation by quadratic formula

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

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

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

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

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

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

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:

First, write the equation in standard form:

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

The quadratic formula is

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

$a=1$, $b=-2$, and $c=-4$. Substituting these values into 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:

First, write the given equation in standard form:

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

The quadratic formula is

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

$a=3$, $b=-5$, and $c=7$. Substituting these values into 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:

Although the given equation is not a quadratic equation, we can convert it into one.

Take, $x^2=y$

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

Substituting these into the given equation,

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

This is a quadratic equation, which 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, we find the $x$ values for each $y$ value.

For $y=10$, substitute it into the equation $x^2=y$:

$x^2=10$

Use the square root property,

$x=\pm \sqrt{10}$

That is, $x=\sqrt{10}$; $x=-\sqrt{10}$

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

$x^2=5$

or

$x=\pm \sqrt{5}$

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

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

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

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:

When you throw a rock upwards, it goes up to a certain height and returns to the ground.

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 becomes zero when the rock reaches the ground.

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

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 $-42t+16t^2$ to both sides, so that we have the quadratic equation in 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 into 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 $t=\dfrac{36}{32}$

Using 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 when the rock reaches the height of $27 ft$ on the way up. The other time, $1.50s$, is when the rock is again at $27 ft$ — this happens on the way down.

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 when it hits the ground, 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 into the quadratic formula,

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

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

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

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

Using 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 $t$,

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

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

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

That is, $t=3.36 s$

Example 1: Use the discriminant to determine the nature of the solutions of $4x^2+12x+9=0$.

Solution: Here $a=4$, $b=12$, and $c=9$. Find the discriminant:

$b^2-4ac=12^2-4\cdot 4\cdot 9=144-144=0$

Since the discriminant is $0$, the equation has one repeated real solution.

Example 2: Use the discriminant to determine the nature of the solutions of $-6x^2+9x-2=0$.

Solution: Here $a=-6$, $b=9$, and $c=-2$. Find the discriminant:

$b^2-4ac=9^2-4\cdot(-6)\cdot(-2)=81-48=33$

The discriminant is positive, so there are two real solutions. Since $33$ is not a perfect square, the equation has two irrational solutions.

Example 3: Use the discriminant to determine the nature of the solutions of $4x^2-x+1=0$.

Solution: Here $a=4$, $b=-1$, and $c=1$. Find the discriminant:

$b^2-4ac=(-1)^2-4\cdot 4\cdot 1=1-16=-15$

Since the discriminant is negative, the equation has two complex solutions that are not real.