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Number line.

• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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• High School Math Solutions – Quadratic Equations Calculator, Part 3 On the last post we covered completing the square (see link). It is pretty strait forward if you follow all the...

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Quadratic Equation Solver

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Step-By-Step Example

Example (click to try), choose your method, solve by factoring.

Example: 3x^2-2x-1=0

Complete The Square

Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.)

Take the Square Root

Example: 2x^2=18

Quadratic Formula

Example: 4x^2-2x-1=0

About quadratic equations

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Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

• Factoring Quadratics
• Completing the Square
• Graphing Quadratic Equations
• The Quadratic Formula
• Online Quadratic Equation Solver

Each example follows three general stages:

• Take the real world description and make some equations
• Use your common sense to interpret the results

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give ac , and add to give b " method in Factoring Quadratics :

ac = −15 , and b = −14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

• t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

• h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

• $700,000 for manufacturing set-up costs, advertising, etc
• $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

• at $0, you just give away 70,000 bikes
• at $350, you won't sell any bikes at all
• at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

• Unit Sales = 70,000 − 200 x 230 = 24,000
• Sales in Dollars = $230 x 24,000 = $5,520,000
• Costs = 700,000 + $110 x 24,000 = $3,340,000
• Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

• Let x = the boat's speed in the water (km/h)
• Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

• when going upstream, v = x−2 (its speed is reduced by 2 km/h)
• when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation!

Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T " is:

1 R T   =   1 R 1 + 1 R 2

In this case, we have R T = 2 and R 2 = R 1 + 3

1 2   =   1 R 1 + 1 R 1 +3

To get rid of the fractions we can multiply all terms by 2R 1 (R 1 + 3) and then simplify:

Yes! A Quadratic Equation!

Let us solve it using our Quadratic Equation Solver .

• Enter 1, −1 and −6
• And you should get the answers −2 and 3

R 1 cannot be negative, so R 1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

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9.3: Solve Quadratic Equations Using the Quadratic Formula

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Learning Objectives

By the end of this section, you will be able to:

• Solve quadratic equations using the quadratic formula
• Use the discriminant to predict the number of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Be Prepared

Before you get started, take this readiness quiz.

• Simplify: \(\frac{−20−5}{10}\). If you missed this problem, review [link] .
• Simplify: \(4+\sqrt{121}\). If you missed this problem, review [link] .
• Simplify: \(\sqrt{128}\). If you missed this problem, review [link] .

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ In this section, we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’ so that we would do the algebraic steps only once and then use the new formula to find the value of the specific variable. Now, we will go through the steps of completing the square in general to solve a quadratic equation for x . It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form\( ax^2+bx+c=0\) as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

This last equation is the Quadratic Formula.

Definition: QUADRATIC FORMULA

The solutions to a quadratic equation of the form \(ax^2+bx+c=0\), \(a\ge 0\) are given by the formula:

\(x=\frac{−b\pm\sqrt{b^2−4ac}}{2a}\)

To use the Quadratic Formula, we substitute the values of a, b, and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

How to Solve a Quadratic Equation Using the Quadratic Formula

Example \(\PageIndex{1}\)

Solve \(2x^2+9x−5=0\) by using the Quadratic Formula.

Example \(\PageIndex{2}\)

Solve \(3y^2−5y+2=0\) by using the Quadratic Formula.

\(y=\frac{2}{3}\), \(y=1\)

Example \(\PageIndex{3}\)

Solve \(4z^2+2z−6=0\) by using the Quadratic Formula.

\(z=−\frac{3}{2}\), \(z=1\) ​​​​​

Definition: SOLVE A QUADRATIC EQUATION USING THE QUADRATIC FORMULA

• Write the Quadratic Formula in standard form. Identify the aa, bb, and cc values.
• Write the Quadratic Formula. Then substitute in the values of a, b, and c.
• Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time. And remember, the Quadratic Formula is an equation. Be sure you start with ‘\(x=\)’.

Example \(\PageIndex{4}\)

Solve \(x^2−6x+5=0\) by using the Quadratic Formula.

Example \(\PageIndex{5}\)

Solve \(a^2−2a−15=0\) by using the Quadratic Formula.

\(a=−3\), \(a=5\)

Example \(\PageIndex{6}\)

Solve \(b^2+10b+24=0\) by using the Quadratic Formula.

\(b=−6\), \(b=−4\)

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Example \(\PageIndex{7}\)

Solve \(4y^2−5y−3=0\) by using the Quadratic Formula.

We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘ x ’.

Example \(\PageIndex{8}\)

Solve \(2p^2+8p+5=0\) by using the Quadratic Formula.

\(p=\frac{−4\pm\sqrt{6}}{2}\)

Example \(\PageIndex{9}\)

Solve \(5q^2−11q+3=0\) by using the Quadratic Formula.

\(q=\frac{11\pm\sqrt{61}}{10}\)

Example \(\PageIndex{10}\)

Solve \(2x^2+10x+11=0\) by using the Quadratic Formula.

Example \(\PageIndex{11}\)

Solve \(3m^2+12m+7=0\) by using the Quadratic Formula.

\(m=\frac{−6\pm\sqrt{15}}{3}\)

Example \(\PageIndex{12}\)

Solve \(5n^2+4n−4=0\) by using the Quadratic Formula.

\(n=\frac{−2\pm2\sqrt{6}}{5}\)

We cannot take the square root of a negative number. So, when we substitute a, b, and c into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

Example \(\PageIndex{13}\)

Solve \(3p^2+2p+9=0\) by using the Quadratic Formula.

Example \(\PageIndex{14}\)

Solve \(4a^2−3a+8=0\) by using the Quadratic Formula.

no real solution

Exeample \(\PageIndex{15}\)

Solve \(5b^2+2b+4=0\) by using the Quadratic Formula.

The quadratic equations we have solved so far in this section were all written in standard form, \(ax^2+bx+c=0\). Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example \(\PageIndex{16}\)

Solve \(x(x+6)+4=0\) by using the Quadratic Formula.

Example \(\PageIndex{17}\)

Solve \(x(x+2)−5=0\) by using the Quadratic Formula.

\(x=−1\pm\sqrt{6}\)

Example \(\PageIndex{18}\)

Solve \(y(3y−1)−2=0\) by using the Quadratic Formula.

\(y=−\frac{2}{3}\), \(y=1\)

When we solved linear equations, if an equation had too many fractions we ‘cleared the fractions’ by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions—to solve. We can use the same strategy with quadratic equations.

Example \(\PageIndex{19}\)

Solve \(\frac{1}{2}u^2+\frac{2}{3}u=\frac{1}{3}\) by using the Quadratic Formula.

Example \(\PageIndex{20}\)

Solve \(\frac{1}{4}c^2−\frac{1}{3}c=\frac{1}{12}\) by using the Quadratic Formula.

\(c=\frac{2\pm\sqrt{7}}{3}\)

Example \(\PageIndex{21}\)

Solve \(\frac{1}{9}d^2−\frac{1}{2}d=−\frac{1}{2}\) by using the Quadratic Formula.

\(d=\frac{3}{2}\), \(d=3\)

Think about the equation \((x−3)^2=0\). We know from the Zero Products Principle that this equation has only one solution: \(x=3\).

We will see in the next example how using the Quadratic Formula to solve an equation with a perfect square also gives just one solution.

Example \(\PageIndex{22}\)

Solve \(4x^2−20x=−25\) by using the Quadratic Formula.

Did you recognize that \(4x^2−20x+25\) is a perfect square?

Example \(\PageIndex{23}\)

Solve \(r^2+10r+25=0\) by using the Quadratic Formula.

\(r=−5\)​​​​​​

Example \(\PageIndex{24}\)

Solve \(25t^2−40t=−16\) by using the Quadratic Formula.

\(t=\frac{4}{5}\)

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant .

Definition: DISCRIMINANT

In the Quadratic Formula\(x=\frac{−b\pm\sqrt{b^2−4ac}}{2a}\), the quantity \(b^2−4ac\) is called the discriminant .

Let’s look at the discriminant of the equations in Example , Example , and Example , and the number of solutions to those quadratic equations.

When the discriminant is positive \(x=\frac{−b\pm\sqrt{+}}{2a}\) the quadratic equation has two solutions .

When the discriminant is zero \(x=\frac{−b\pm\sqrt{0}}{2a}\) the quadratic equation has one solution .

When the discriminant is negative \(x=\frac{−b\pm\sqrt{−}}{2a}\) the quadratic equation has no real solutions .

Definition:USE THE DISCRIMINANT, \(b^2−4ac\), TO DETERMINE THE NUMBER OF SolutionS OF A QUADRATIC EQUATION

For a quadratic equation of the form \(ax^2+bx+c=0\), \(a \ge 0\),

• if \(b^2−4ac>0\), the equation has two solutions.
• if \(b^2−4ac=0\), the equation has one solution.
• if \(b^2−4ac<0\), the equation has no real solutions.

Example \(\PageIndex{25}\)

Determine the number of solutions to each quadratic equation:

• \(2v^2−3v+6=0\)
• \(3x^2+7x−9=0\)
• \(5n^2+n+4=0\)
• \(9y^2−6y+1=0\)

Example \(\PageIndex{26}\)

• \(8m^2−3m+6=0\)
• \(5z^2+6z−2=0\)
• \(9w^2+24w+16=0\)
• \(9u^2−2u+4=0\)
• no real solutions

Example \(\PageIndex{27}\)

​​​​​​​Determine the number of solutions to each quadratic equation:

• \( b^2+7b−13=0\)
• \(5a^2−6a+10=0\)
• \(4r^2−20r+25=0\)
• \(7t^2−11t+3=0\)

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We have used four methods to solve quadratic equations:

• Square Root Property
• Completing the Square
• Quadratic Formula

You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use.

Definition: IDENTIFY THE MOST APPROPRIATE METHOD TO SOLVE A QUADRATIC EQUATION.

• Try Factoring first. If the quadratic factors easily, this method is very quick.
• Try the Square Root Property next. If the equation fits the form \(ax^2=k\) or \(a(x−h)^2=k\), it can easily be solved by using the Square Root Property.
• Use the Quadratic Formula . Any quadratic equation can be solved by using the Quadratic Formula.

​​​​​​What about the method of completing the square? Most people find that method cumbersome and prefer not to use it. We needed to include it in this chapter because we completed the square in general to derive the Quadratic Formula. You will also use the process of completing the square in other areas of algebra.

Example \(\PageIndex{28}\)

Identify the most appropriate method to use to solve each quadratic equation:

• \(5z^2=17\)
• \(4x^2−12x+9=0\)
• \(8u^2+6u=11\)

1. \(5z^2=17\)

Since the equation is in the \(ax^2=k\), the most appropriate method is to use the Square Root Property.

2. \(4x^2−12x+9=0\)

We recognize that the left side of the equation is a perfect square trinomial, and so Factoring will be the most appropriate method.

3. \(8u^2+6u=11\)

Put the equation in standard form. \(8u^2+6u−11=0\)

While our first thought may be to try Factoring, thinking about all the possibilities for trial and error leads us to choose the Quadratic Formula as the most appropriate method.

Example \(\PageIndex{29}\)

• \(x^2+6x+8=0\)
• \((n−3)^2=16\)
• \(5p^2−6p=9\)

Example \(\PageIndex{30}\)

• \(8a^2+3a−9=0\)
• \(4b^2+4b+1=0\)
• \(5c2=125\)

​​​​​​​ Access these online resources for additional instruction and practice with using the Quadratic Formula:

• Solving Quadratic Equations: Solving with the Quadratic Formula
• How to solve a quadratic equation in standard form using the Quadratic Formula (example)
• Solving Quadratic Equations using the Quadratic Formula—Example 3
• Solve Quadratic Equations using Quadratic Formula

Key Concepts

• Write the quadratic formula in standard form. Identify the a, b, c values.
• Write the quadratic formula. Then substitute in the values of a, b, c.
• if \(b^2−4ac>0\), the equation has 2 solutions.
• if \(b^2−4ac=0\), the equation has 1 solution.
• Try Factoring first. If the quadratic factors easily this method is very quick.
• Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

Quadratic Formula Calculator

Enter the equation you want to solve using the quadratic formula.

The Quadratic Formula Calculator finds solutions to quadratic equations with real coefficients. For equations with real solutions, you can use the graphing tool to visualize the solutions.

Quadratic Formula : x = − b ± b 2 − 4 a c 2 a

Click the blue arrow to submit. Choose "Solve Using the Quadratic Formula" from the topic selector and click to see the result in our Algebra Calculator !

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Course: Algebra 1   >   Unit 14

• Solve by completing the square: Integer solutions
• Solve by completing the square: Non-integer solutions
• Solve equations by completing the square
• Worked example: completing the square (leading coefficient ≠ 1)
• Completing the square
• Solving quadratics by completing the square: no solution
• Proof of the quadratic formula

Solving quadratics by completing the square

• Completing the square review
• Quadratic formula proof review

What you should be familiar with before taking this lesson

• Solving quadratic equations by taking the square root
• Solving quadratic equations by factoring

What you will learn in this lesson

Solving quadratic equations by completing the square, what happened here, how to complete the square.

• The coefficient of x ‍   , which we know to be 6 ‍   , should be equal to 2 a ‍   . This means that a = 3 ‍   .
• The constant number we need to add is equal to a 2 ‍   , which is 3 2 = 9 ‍   .
• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
• (Choice A)   b 2 ‍   A b 2 ‍  
• (Choice B)   ( b 2 ) 2 ‍   B ( b 2 ) 2 ‍  
• (Choice C)   ( 2 b ) 2 ‍   C ( 2 b ) 2 ‍  
• (Choice D)   b 2 ‍   D b 2 ‍  

Solving equations one more time

• (Choice A)   x = 21 − 4 ‍   and − 21 − 4 ‍   A x = 21 − 4 ‍   and − 21 − 4 ‍  
• (Choice B)   x = 69 − 8 ‍   and − 69 − 8 ‍   B x = 69 − 8 ‍   and − 69 − 8 ‍  
• (Choice C)   x = 21 + 4 ‍   and − 21 + 4 ‍   C x = 21 + 4 ‍   and − 21 + 4 ‍  
• (Choice D)   x = 69 + 8 ‍   and − 69 + 8 ‍   D x = 69 + 8 ‍   and − 69 + 8 ‍  
• (Choice A)   x = 1 2 + 3 4 ‍   and − 1 2 + 3 4 ‍   A x = 1 2 + 3 4 ‍   and − 1 2 + 3 4 ‍  
• (Choice B)   x = 2 + 3 2 ‍   and − 2 + 3 2 ‍   B x = 2 + 3 2 ‍   and − 2 + 3 2 ‍  
• (Choice C)   x = 1 2 − 3 4 ‍   and − 1 2 − 3 4 ‍   C x = 1 2 − 3 4 ‍   and − 1 2 − 3 4 ‍  
• (Choice D)   x = 2 − 3 2 ‍   and − 2 − 3 2 ‍   D x = 2 − 3 2 ‍   and − 2 − 3 2 ‍  

Arranging the equation before completing the square

Rule 1: separate the variable terms from the constant term, rule 2: make sure the coefficient of x 2 ‍   is equal to 1 ‍   ..

• (Choice A)   x = 7 − 5 2 ‍   and − 7 − 5 2 ‍   A x = 7 − 5 2 ‍   and − 7 − 5 2 ‍  
• (Choice B)   x = 10 − 5 ‍   and − 10 − 5 ‍   B x = 10 − 5 ‍   and − 10 − 5 ‍  
• (Choice C)   x = 7 + 5 2 ‍   and − 7 + 5 2 ‍   C x = 7 + 5 2 ‍   and − 7 + 5 2 ‍  
• (Choice D)   x = 10 + 5 ‍   and − 10 + 5 ‍   D x = 10 + 5 ‍   and − 10 + 5 ‍  

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