# 2.5 Quadratic equations  (Page 6/14)

 Page 6 / 14

## Finding the length of the missing side of a right triangle

Find the length of the missing side of the right triangle in [link] .

As we have measurements for side b and the hypotenuse, the missing side is a.

$\begin{array}{ccc}\hfill {a}^{2}+{b}^{2}& =& {c}^{2}\hfill \\ {a}^{2}+{\left(4\right)}^{2}\hfill & =& {\left(12\right)}^{2}\hfill \\ \hfill {a}^{2}+16& =& 144\hfill \\ \hfill {a}^{2}& =& 128\hfill \\ \hfill a& =& \sqrt{128}\hfill \\ & =& 8\sqrt{2}\hfill \end{array}$

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

$5\text{\hspace{0.17em}}$ units

## Key equations

 quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

## Key concepts

• Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions. See [link] , [link] , and [link] .
• Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. See [link] and [link] .
• Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. See [link] and [link] .
• Completing the square is a method of solving quadratic equations when the equation cannot be factored. See [link] .
• A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See [link] .
• The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See [link] .
• The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See [link] .

## Verbal

How do we recognize when an equation is quadratic?

It is a second-degree equation (the highest variable exponent is 2).

When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0\text{\hspace{0.17em}}$ we may graph the equation $\text{\hspace{0.17em}}y=a{x}^{2}+bx+c\text{\hspace{0.17em}}$ and have no zeroes ( x -intercepts).

When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

We want to take advantage of the zero property of multiplication in the fact that if $\text{\hspace{0.17em}}a\cdot b=0\text{\hspace{0.17em}}$ then it must follow that each factor separately offers a solution to the product being zero:

In the quadratic formula, what is the name of the expression under the radical sign $\text{\hspace{0.17em}}{b}^{2}-4ac,$ and how does it determine the number of and nature of our solutions?

Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

One, when no linear term is present (no x term), such as $\text{\hspace{0.17em}}{x}^{2}=16.\text{\hspace{0.17em}}$ Two, when the equation is already in the form $\text{\hspace{0.17em}}{\left(ax+b\right)}^{2}=d.$

Cos45/sec30+cosec30=
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1 By By   By  By Danielrosenberger  By By Nick Swain By Jordon Humphreys 