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## Section 2.5 : Quadratic Equations - Part I

For problems 1 – 7 solve the quadratic equation by factoring.

- \({u^2} - 5u - 14 = 0\) Solution
- \({x^2} + 15x = - 50\) Solution
- \({y^2} = 11y - 28\) Solution
- \(19x = 7 - 6{x^2}\) Solution
- \(6{w^2} - w = 5\) Solution
- \({z^2} - 16z + 61 = 2z - 20\) Solution
- \(12{x^2} = 25x\) Solution

For problems 8 & 9 use factoring to solve the equation.

For problems 10 – 12 use factoring to solve the equation.

- \(\displaystyle \frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\) Solution
- \(\displaystyle \frac{{4z}}{{z + 1}} + \frac{5}{z} = \frac{{6z + 5}}{{{z^2} + z}}\) Solution
- \(\displaystyle x + 1 = \frac{{2x - 7}}{{x + 5}} - \frac{{5x + 8}}{{x + 5}}\) Solution

For problems 13 – 16 use the Square Root Property to solve the equation.

- \(9{u^2} - 16 = 0\) Solution
- \({x^2} + 15 = 0\) Solution
- \({\left( {z - 2} \right)^2} - 36 = 0\) Solution
- \({\left( {6t + 1} \right)^2} + 3 = 0\) Solution

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A quadratic equation is an equation that could be written as

To solve a quadratic equation by factoring,

- Put all terms on one side of the equal sign, leaving zero on the other side.
- Set each factor equal to zero.
- Solve each of these equations.
- Check by inserting your answer in the original equation.

x 2 – 6 x = 16 becomes x 2 – 6 x – 16 = 0

Both values, 8 and –2, are solutions to the original equation.

Setting all terms equal to zero,

Solve 2 x 2 + 2 x – 1 = x 2 + 6 x – 5.

First, simplify by putting all terms on one side and combining like terms.

To check, 2 x 2 + 2 x – 1 = x 2 + 6 x – 5

a, b, and c are taken from the quadratic equation written in its general form of

- Two different real roots if the discriminant b 2 – 4 ac is a positive number.
- One real root if the discriminant b 2 – 4 ac is equal to 0.
- No real root if the discriminant b 2 – 4 ac is a negative number.

Because the discriminant b 2 – 4 ac is positive, you get two different real roots.

Then substitute 1, 2, and –2 for a , b , and c, respectively, in the quadratic formula and simplify.

Note that the two roots are irrational.

Solve for x : x 2 + 2 x + 1 = 0.

Substituting in the quadratic formula,

Since the discriminant b 2 – 4 ac is 0, the equation has one root.

Solve for x : x ( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0.

Solve for x : x 2 – 6 x + 5 = 0.

Take the square root of both sides.

Solve for y : y 2 + 2 y – 4 = 0.

Solve for x : 2 x 2 + 3 x + 2 = 0.

Previous Quiz: Operations with Square Roots

Next Quiz: Solving Quadratic Equations

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Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

Hence, by the condition of the problem

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore, Perimeter = 2(2 + 5) m = 14 m.

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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## Quadratic Equations

An example of a Quadratic Equation :

The function makes nice curves like this one:

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2 ).

It is also called an "Equation of Degree 2" (because of the "2" on the x )

## Standard Form

The Standard Form of a Quadratic Equation looks like this:

## Have a Play With It

Play with the " Quadratic Equation Explorer " so you can see:

## Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

But sometimes a quadratic equation does not look like that!

## How To Solve Them?

The " solutions " to the Quadratic Equation are where it is equal to zero .

They are also called " roots ", or sometimes " zeros "

There are usually 2 solutions (as shown in this graph).

And there are a few different ways to find the solutions:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

## About the Quadratic Formula

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

Here is an example with two answers:

But it does not always work out like that!

- Imagine if the curve "just touches" the x-axis.
- Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

## Discriminant

- when b 2 − 4ac is positive, we get two Real solutions
- when it is zero we get just ONE real solution (both answers are the same)
- when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

## Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

## Example: Solve 5x 2 + 6x + 1 = 0

And we see them on this graph.

## Remembering The Formula

A kind reader suggested singing it to "Pop Goes the Weasel":

Try singing it a few times and it will get stuck in your head!

Or you can remember this story:

## Complex Solutions?

It means our answer will include Imaginary Numbers . Wow!

## Example: Solve 5x 2 + 2x + 1 = 0

√(−16) = 4 i (where i is the imaginary number √−1)

The graph does not cross the x-axis. That is why we ended up with complex numbers.

In a way it is easier: we don't need more calculation, we leave it as −0.2 ± 0.4 i .

## Example: Solve x 2 − 4x + 6.25 = 0

√(−9) = 3 i (where i is the imaginary number √−1)

Just an interesting fact for you!

- Quadratic Equation in Standard Form: ax 2 + bx + c = 0
- Quadratic Equations can be factored
- Quadratic Formula: x = −b ± √(b 2 − 4ac) 2a
- positive, there are 2 real solutions
- zero, there is one real solution
- negative, there are 2 complex solutions

## Quadratic Equation Questions

Definition of Quadratic Equation

There are different methods to find the roots of quadratic equation, such as:

Learn: Factorization of Quadratic equations

The quadratic formula to find the roots of the quadratic equation is given by:

\(\begin{array}{l}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\end{array} \)

Where b 2 -4ac is called the discriminant of the equation.

- two distinct real roots, if b 2 – 4ac > 0
- two equal real roots, if b 2 – 4ac = 0
- no real roots, if b 2 – 4ac < 0

Also, learn quadratic equations for class 10 here.

## Quadratic Equations Problems and Solutions

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

This represents the quadratic equation. Hence by solving the given equation for x, we get;

So, the number of marbles Rahul had is 36 and Rohan had is 9 or vice versa.

2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

x(x + 1) + 8 = (x + 2) (x – 2)

Since, this expression is not in the form of ax 2 +bx+c, hence it is not a quadratic equation.

3. Find the roots of the equation 2x 2 – 5x + 3 = 0 using factorisation.

Therefore, 3/2 and 1 are the roots of the given equation.

4. Solve the quadratic equation 2x 2 + x – 300 = 0 using factorisation.

Therefore, 12 and -12.5 are two roots of the given equation.

5. Solve the equation x 2 +4x-5=0.

6. Solve the quadratic equation 2x 2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax 2 +bx+c = 0

Hence, by using the quadratic formula:

Now putting the values of a,b and c.

7. Find the roots of x 2 + 4x + 5 = 0, if any exist, using quadratic formula.

D = b 2 -4ac = 4 2 – 4(1)(5) = 16-20 = -4

8. Find the discriminant of the equation: 3x 2 -2x+⅓ = 0.

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b 2 – 4ac

## Video Lesson

## Practice Questions

Solve these quadratic equations and find the roots.

- x 2 -5x-14=0 [Answer: x=-2 & x=7]
- X 2 = 11x -28 [Answer: x=4 & x = 7]
- 6x 2 – x = 5 [Answer: x=-⅚ & x = 1]
- 12x 2 = 25x [Answer: x=0 & x=25/12]

## Frequently Asked Questions on Quadratic Equations

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Section 2.5 : Quadratic Equations - Part I · w2−10w+2+w−4=w−3 w 2 − 10 w + 2 + w − 4 = w − 3 Solution · 4zz+1+5z=6z+5z2+z 4 z z + 1 + 5 z =

Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a

Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a

Solving Quadratic Equations · Put the equation into the form ax 2 + bx = – c. · Make sure that a = 1 (if a ≠ 1, multiply through the equation by before

TabletClass Math:https://tcmathacademy.com/ Quadratic equations practice problems and solutions. For more math help to include math lessons

Factoring and Solving Quadratic Equations Worksheet. Math Tutorial Lab Special Topic∗. Example Problems. Factor completely. 1. 3x + 36. 2. 4x2 + 16x.

1. Solve using the quadratic formula: x2 - 2x - 24 = 0. spy1.

Solving Quadratic Equations · Now we will learn how to frame the equations from word problem: · 1. · Solution: · Method I: Using only one unknown · Let the two

Example: Solve 5x2 + 6x + 1 = 0 ; Quadratic Formula: · −b ± √(b2 − 4ac) 2a ; Put in a, b and c: · −6 ± √(62 − 4×5×1) 2×5 ; Solve: · −6 ± √(36− 20) 10 ; −6 ±

Practice Questions · x2-5x-14=0 [Answer: x=-2 & x=7] · X2 = 11x -28 [Answer: x=4 & x = 7] · 6x2– x = 5 [Answer: x=-⅚ & x = 1] · 12x2 = 25x [Answer: