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Lesson Explainer: Solving Quadratic Equations: Quadratic Formula Mathematics • First Year of Secondary School

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In this explainer, we will learn how to solve quadratic equations using the quadratic formula.

Recall that a quadratic equation is an equation with one variable, where the highest order of any term is 2. This is made more explicit in the definition below.

Definition: Quadratic Equation

The equation ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, with variable ๐‘ฅ and constants ๐‘Ž, ๐‘, and ๐‘, is called a quadratic equation (or second-degree equation).

To solve quadratic equations, we can use the following methods:

  • Factoring
  • Completing the square
  • Using the quadratic formula
  • Solving graphically

So far, we have met factoring and completing the square. We will use completing the square to derive the quadratic formula, which is the method we will be focusing on in this explainer.

If we have some quadratic equation of the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, with variable ๐‘ฅ and constants ๐‘Ž, ๐‘, and ๐‘, then using completing the square, we can rearrange and solve for ๐‘ฅ as follows.

First, we will divide through by ๐‘Ž, the coefficient of ๐‘ฅ๏Šจ: ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๐‘ฅ+๐‘๐‘Ž๐‘ฅ+๐‘๐‘Ž=0.๏Šจ๏Šจ

Next, we will subtract ๐‘๐‘Ž from both sides: ๐‘ฅ+๐‘๐‘Ž๐‘ฅ=โˆ’๐‘๐‘Ž.๏Šจ

Then, to make a perfect square, we will add ๏€ฝ๐‘2๐‘Ž๏‰๏Šจ to both sides: ๐‘ฅ+๐‘๐‘Ž๐‘ฅ+๏€ฝ๐‘2๐‘Ž๏‰=โˆ’๐‘๐‘Ž+๏€ฝ๐‘2๐‘Ž๏‰.๏Šจ๏Šจ๏Šจ

Since the left-hand side of the equation is in the form ๐‘ฅ+2๐‘‘๐‘ฅ+๐‘‘๏Šจ๏Šจ, a perfect square, then we can write this as (๐‘ฅ+๐‘‘)๏Šจ: ๏€ฝ๐‘ฅ+๐‘2๐‘Ž๏‰=โˆ’๐‘๐‘Ž+๏€ฝ๐‘2๐‘Ž๏‰.๏Šจ๏Šจ

Next, we will rearrange to get ๐‘ฅ as the subject. First, we square root both sides (taking both the positive and negative square roots): ๏€ฝ๐‘ฅ+๐‘2๐‘Ž๏‰=โˆ’๐‘๐‘Ž+๏€ฝ๐‘2๐‘Ž๏‰๐‘ฅ+๐‘2๐‘Ž=ยฑ๏„Ÿโˆ’๐‘๐‘Ž+๏€ฝ๐‘2๐‘Ž๏‰.๏Šจ๏Šจ๏Šจ

Next, we isolate ๐‘ฅ: ๐‘ฅ=โˆ’๐‘2๐‘Žยฑ๏„Ÿโˆ’๐‘๐‘Ž+๏€ฝ๐‘2๐‘Ž๏‰.๏Šจ

We then simplify further by expanding the brackets in the radical and rearranging slightly: ๐‘ฅ=โˆ’๐‘2๐‘Žยฑ๏„Ÿ๏€ฝ๐‘2๐‘Ž๏‰โˆ’๐‘๐‘Ž=โˆ’๐‘2๐‘Žยฑ๏„ž๐‘4๐‘Žโˆ’๐‘๐‘Ž.๏Šจ๏Šจ๏Šจ

We can now write the whole expression in the radical over a common denominator: ๐‘ฅ=โˆ’๐‘2๐‘Žยฑ๏„ž๐‘4๐‘Žโˆ’4๐‘Ž๐‘4๐‘Ž=โˆ’๐‘2๐‘Žยฑ๏„ž๐‘โˆ’4๐‘Ž๐‘4๐‘Ž.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Since the denominator in the radical is a perfect square, then we can square root this and write it outside of the radical: ๐‘ฅ=โˆ’๐‘2๐‘Žยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

Combining the two terms, since they have the same denominator, we then get ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

So in its final form, ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž๏Šจ is the quadratic formula, which is used for finding solutions of quadratic equations of the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0. This is stated in the definition below.

Definition: The Quadratic Formula

To solve a quadratic equation in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, with variable ๐‘ฅ and constants ๐‘Ž, ๐‘, and ๐‘ we can use the quadratic formula to solve for ๐‘ฅ: ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

Whenever we want to use the quadratic formula, we need to ensure that our quadratic equation is equal to zero, with it in expanded form and simplified as much as possible, so that it is in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0. Then, we need to identify what ๐‘Ž, ๐‘, and ๐‘ each are. Following that, we can substitute into the quadratic formula and solve for ๐‘ฅ. There are usually two values of ๐‘ฅ, corresponding to the positive and negative square roots of ๐‘โˆ’4๐‘Ž๐‘๏Šจ, but sometimes there is only one, or even no real solutions, depending on the values of ๐‘Ž, ๐‘, and ๐‘.

We will discuss how to solve a quadratic equation which is already in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0 in our first example.

Example 1: Solving Quadratic Equations Using the Quadratic Formula

Find the solution set of the equation 6๐‘ฅโˆ’8๐‘ฅ+1=0๏Šจ, giving values to two decimal places.

Answer

As the equation 6๐‘ฅโˆ’8๐‘ฅ+1=0๏Šจ is a quadratic equation, then we can use one of the methods for solving quadratics. In this case, we are going to use the quadratic formula in order to solve. Recall, that for a quadratic equation in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, then ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

Since 6๐‘ฅโˆ’8๐‘ฅ+1=0๏Šจ is already in the same form as ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ since it is equal to zero, fully simplified, and written in descending powers of ๐‘ฅ, then we can identify the values of ๐‘Ž, ๐‘, and ๐‘: ๐‘Ž=6,๐‘=โˆ’8,๐‘=1.and

Substituting these values into the quadratic formula and solving for ๐‘ฅ, we get ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž=โˆ’(โˆ’8)ยฑ๏„(โˆ’8)โˆ’4(6)(1)2(6)=8ยฑโˆš64โˆ’2412=8ยฑโˆš4012=8ยฑ2โˆš1012=4ยฑโˆš106.๏Šจ๏Šจ

As we are asked to give the solution set to 2 decimal places, then evaluating ๐‘ฅ, we get ๐‘ฅ=1.19โ€ฆ0.14โ€ฆ.or

So, the solution set of the equation 6๐‘ฅโˆ’8๐‘ฅ+1=0๏Šจ is {0.14,1.19} correct to 2 decimal places.

In the next example, we will consider how to solve a quadratic equation that is not in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ initially but, by rearranging, can be put in this form and then solved using the quadratic formula.

Example 2: Solving Quadratic Equations Using the Quadratic Formula

Find the solution set of ๐‘ฅโˆ’6(๐‘ฅโˆ’1)=2๏Šจ in โ„, giving values to two decimal places.

Answer

As the equation ๐‘ฅโˆ’6(๐‘ฅโˆ’1)=2๏Šจ contains a term with ๐‘ฅ๏Šจ, then this is likely to be a quadratic equation. To check, we can simplify by expanding brackets and making the equation equal to zero, as follows: ๐‘ฅโˆ’6(๐‘ฅโˆ’1)=2๐‘ฅโˆ’6๐‘ฅ+6=2๐‘ฅโˆ’6๐‘ฅ+4=0.๏Šจ๏Šจ๏Šจ

As the highest power in the equation is 2, then we can see this is a quadratic equation. Since it is now written in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0 then we can apply the quadratic formula, which states ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

We can see that for ๐‘ฅโˆ’6๐‘ฅ+4=0๏Šจ, ๐‘Ž=1, ๐‘=โˆ’6, and ๐‘=4. Substituting this into the quadratic formula, we get ๐‘ฅ=โˆ’(โˆ’6)ยฑ๏„(โˆ’6)โˆ’4(1)(4)2(1).๏Šจ

Simplifying, we get ๐‘ฅ=6ยฑโˆš36โˆ’162=6ยฑโˆš202=6ยฑ2โˆš52=3ยฑโˆš5.

As we are asked to give the solution set to 2 decimal places, then evaluating ๐‘ฅ, we get ๐‘ฅ=5.24โ€ฆ๐‘ฅ=0.76โ€ฆ.or

So, the solution set of ๐‘ฅโˆ’6(๐‘ฅโˆ’1)=2๏Šจ correct to 2 decimal places is {0.76,5.24}.

We can use the solutions of quadratic equations to find unknown parts of the equation, such as coefficients or constants. We can do this by substituting known and unknown parts into the quadratic formula and solving for the unknown part. We will explore how to do this in the next example.

Example 3: Finding Unknowns in Quadratic Equations Using the Quadratic Formula

Given that ๐‘ฅ=โˆ’2 is a root of the equation ๐‘ฅโˆ’4๐‘š๐‘ฅโˆ’๏€น๐‘šโˆ’6๏…=0๏Šจ๏Šจ, find the set of possible values of ๐‘š.

Answer

Since ๐‘ฅโˆ’4๐‘š๐‘ฅโˆ’๏€น๐‘šโˆ’6๏…=0๏Šจ๏Šจ is of the form of a quadratic equation, then we can use the quadratic formula to find the values of ๐‘š.

The quadratic formula states that for an equation in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0 then ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

We can see by comparing ๐‘ฅโˆ’4๐‘š๐‘ฅโˆ’๏€น๐‘šโˆ’6๏…=0๏Šจ๏Šจ with ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ that ๐‘Ž=1, ๐‘=โˆ’4๐‘š, and ๐‘=โˆ’๏€น๐‘šโˆ’6๏…๏Šจ. Since we also know that one of the roots of the equation is ๐‘ฅ=โˆ’2 then we can substitute ๐‘Ž, ๐‘, ๐‘ and ๐‘ฅ into the quadratic formula: ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Žโˆ’2=โˆ’(โˆ’4๐‘š)ยฑ๏„(โˆ’4๐‘š)โˆ’4(โˆ’(๐‘šโˆ’6)(1))2(1).๏Šจ๏Šจ๏Šจ

Simplifying, we get โˆ’2=4๐‘šยฑโˆš16๐‘š+4(๐‘šโˆ’6)2=4๐‘šยฑโˆš16๐‘š+4๐‘šโˆ’242=4๐‘šยฑโˆš20๐‘šโˆ’242.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

We can simplify further by factoring out 4 and putting it outside the radical: โˆ’2=4๐‘šยฑโˆš4(5๐‘šโˆ’6)2=4๐‘šยฑ2โˆš(5๐‘šโˆ’6)2=2๐‘šยฑโˆš(5๐‘šโˆ’6).๏Šจ๏Šจ๏Šจ

Next, we need to solve for ๐‘š. Since part of the equation contains ๐‘š๏Šจ, then once rearranged, it is likely to be the form of a quadratic equation. As such, we want to rearrange so itโ€™s in the form ๐‘Ž๐‘š+๐‘๐‘š+๐‘=0๏Šจ, ๐‘Žโ‰ 0, so that we can apply the quadratic formula again (but this time with different values for ๐‘Ž, ๐‘, and ๐‘).

Rearranging, we get โˆ’2=2๐‘šยฑโˆš(5๐‘šโˆ’6)โˆ’2โˆ’2๐‘š=ยฑโˆš(5๐‘šโˆ’6)(โˆ’2โˆ’2๐‘š)=5๐‘šโˆ’64+8๐‘š+4๐‘š=5๐‘šโˆ’60=5๐‘šโˆ’4๐‘šโˆ’8๐‘šโˆ’4โˆ’6๐‘šโˆ’8๐‘šโˆ’10=0.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Now that the equation is in the form ๐‘Ž๐‘š+๐‘๐‘š+๐‘=0๏Šจ, ๐‘Žโ‰ 0, we can find ๐‘Ž, ๐‘, and ๐‘. By comparing, we can see that ๐‘Ž=1, ๐‘=โˆ’8, and ๐‘=โˆ’10. Substituting into the quadratic formula, we get ๐‘š=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž=โˆ’(โˆ’8)ยฑ๏„(โˆ’8)โˆ’4(1)(โˆ’10)2(1).๏Šจ๏Šจ

Simplifying, we get ๐‘š=8ยฑโˆš64+402=8ยฑโˆš1042=8ยฑ2โˆš262=4ยฑโˆš26.

Therefore, the possible values of ๐‘š are {4โˆ’โˆš26,4+โˆš26}.

In addition to quadratic equations that contain quadratic terms, we can have equations that may not appear to be quadratic on first inspection but with some rearranging become quadratics. For example, 1๐‘ฅ=4๐‘ฅ can be rearranged to give 1=4๐‘ฅ๏Šจ, which is a quadratic equation. Therefore, we can solve equations, that once rearranged, become quadratics, and can do so using the quadratic formula. In our next example we will explore how to do this.

Example 4: Rearranging Equation to Solve Using the Quadratic Formula

Find the solution set of the equation โˆ’5โˆ’5๐‘ฅ=1๐‘ฅ๏Šจ in โ„, giving values to one decimal place.

Answer

In order to solve โˆ’5โˆ’5๐‘ฅ=1๐‘ฅ๏Šจ, it is helpful to remove any variables in the denominators first. To do this we need to multiply through by the highest power of ๐‘ฅ in the denominator, which is ๐‘ฅ๏Šจ. Doing so gives us โˆ’5โˆ’5๐‘ฅ=1๐‘ฅโˆ’5๐‘ฅโˆ’5๐‘ฅ๐‘ฅ=๐‘ฅ๐‘ฅโˆ’5๐‘ฅโˆ’5๐‘ฅ=1.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

As the highest power of the equation is 2, then we can see this is a quadratic equation. To solve using the quadratic formula, we need to rearrange to make this in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0. In this case, it is helpful to move the terms on the left-hand side to the right-hand side so that the coefficients are positive (but it does not necessarily matter) so that it is easier to do calculations later: โˆ’5๐‘ฅโˆ’5๐‘ฅ=10=5๐‘ฅ+5๐‘ฅ+15๐‘ฅ+5๐‘ฅ+1=0.๏Šจ๏Šจ๏Šจ

Now that this is in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0 we can apply the quadratic formula, which states ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

Comparing 5๐‘ฅ+5๐‘ฅ+1=0๏Šจ with ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, we can see that ๐‘Ž=5, ๐‘=5, and ๐‘=1. Substituting into the quadratic formula, we get ๐‘ฅ=โˆ’5ยฑโˆš5โˆ’4(5)(1)2(5).๏Šจ

Simplifying, we get ๐‘ฅ=โˆ’5ยฑโˆš25โˆ’2010=โˆ’5ยฑโˆš510.

As the question requires us to find the solution set accurate to one decimal place, then we will evaluate this, giving us ๐‘ฅ=โˆ’0.28โ€ฆ๐‘ฅ=โˆ’0.72.or

Therefore, the solution set for the equation โˆ’5โˆ’5๐‘ฅ=1๐‘ฅ๏Šจ correct to one decimal place is {โˆ’0.3,โˆ’0.7}.

In addition to equations that can be rearranged to give a quadratic equation, we can have equations that are not quadratics themselves but can be solved using the quadratic formula as they are in a quadratic form. For example, ๐‘ฅ+2๐‘ฅ+1=0๏Šช๏Šจ is a quartic equation, as its highest power is 4, but as it takes the form of ๐‘Ž๐‘ฆ+๐‘๐‘ฆ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, where the variable ๐‘ฆ represents ๐‘ฅ๏Šจ or ๏€น๐‘ฅ๏…+2๏€น๐‘ฅ๏…+1=0๏Šจ๏Šจ๏Šจ, then it can be solved using the quadratic formula since it has a quadratic form.

In the next example, we will consider how to solve a quartic equation by writing it in the form of a quadratic equation and using the quadratic formula.

Example 5: Using the Quadratic Formula to Solve a Quartic Equation

Using the quadratic formula, find all the solutions to ๐‘ฅโˆ’10๐‘ฅ+1=0๏Šช๏Šจ.

Answer

To find the solutions of the equation ๐‘ฅโˆ’10๐‘ฅ+1=0๏Šช๏Šจ using the quadratic formula, we need to write the equation in the form of a quadratic equation. We can see that we have even powers of ๐‘ฅ, meaning we can replace ๐‘ฅ๏Šจ with another variable, say ๐‘ฆ, giving us ๐‘ฆ=๐‘ฅ๏Šจ. Doing so gives us ๐‘ฅโˆ’10๐‘ฅ+1=0๐‘ฆโˆ’10๐‘ฆ+1=0๏Šช๏Šจ๏Šจ

We can now see that ๐‘ฆโˆ’10๐‘ฆ+1=0๏Šจ takes the form of a quadratic equation ๐‘Ž๐‘ฆ+๐‘๐‘ฆ+๐‘=0๏Šจ, ๐‘Žโ‰ 0. We can then apply the quadratic formula to solve for ๐‘ฆ, which states ๐‘ฆ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ

Since ๐‘Ž=1, ๐‘=โˆ’10, and ๐‘=1, then substituting gives us ๐‘ฆ=โˆ’(โˆ’10)ยฑ๏„(โˆ’10)โˆ’4(1)(1)2(1).๏Šจ

Simplifying, we then get ๐‘ฆ=10ยฑโˆš100โˆ’42=10ยฑโˆš962=10ยฑ4โˆš62=5ยฑ2โˆš6.

Remember that we let ๐‘ฆ=๐‘ฅ๏Šจ, so ๐‘ฅ=5ยฑ2โˆš6.๏Šจ

Square rooting both sides and solving for ๐‘ฅ gives us ๐‘ฅ=ยฑ๏„5ยฑ2โˆš6.

Therefore, all solutions to the equation ๐‘ฅโˆ’10๐‘ฅ+1=0๏Šช๏Šจ are ๐‘ฅ=๏„5+2โˆš6,๐‘ฅ=โˆ’๏„5+2โˆš6,๐‘ฅ=๏„5โˆ’2โˆš6,๐‘ฅ=โˆ’๏„5โˆ’2โˆš6.and

In this explainer, we have discussed how to solve quadratic equations using the quadratic formula. In order to get equations into the form required, we have either rearranged the equation, or substituted to make it into a quadratic form. Letโ€™s recap the key points.

Key Points

  • We can solve quadratic equations of the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0 using the quadratic formula ๐‘ฅ=โˆ’๐‘ยฑโˆš๐‘โˆ’4๐‘Ž๐‘2๐‘Ž.๏Šจ
  • By rearranging, some equations can be written as a quadratic equations of the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, and then solved using the quadratic formula.
  • Some equations that are not quadratic equations themselves, can be solved using the quadratic formula if they take the form ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘=0๏Šจ, ๐‘Žโ‰ 0, where ๐‘ฅ can represent another function.

ุงู†ุถู… ุฅู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ

ุดุงุฑูƒ ููŠ ุงู„ุญุตุต ุงู„ู…ุจุงุดุฑุฉ ุนู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ ูˆุญู‚ู‚ ุงู„ุชู…ูŠุฒ ุงู„ุฏุฑุงุณูŠ ุจุฅุฑุดุงุฏ ูˆุชูˆุฌูŠู‡ ู…ู† ู…ุฏุฑุณ ุฎุจูŠุฑ!

  • ุญุตุต ุชูุงุนู„ูŠุฉ
  • ุฏุฑุฏุดุฉ ูˆุฑุณุงุฆู„
  • ุฃุณุฆู„ุฉ ุงู…ุชุญุงู†ุงุช ูˆุงู‚ุนูŠุฉ

ุชุณุชุฎุฏู… «ู†ุฌูˆู‰» ู…ู„ูุงุช ุชุนุฑูŠู ุงู„ุงุฑุชุจุงุท ู„ุถู…ุงู† ุญุตูˆู„ูƒ ุนู„ู‰ ุฃูุถู„ ุชุฌุฑุจุฉ ุนู„ู‰ ู…ูˆู‚ุนู†ุง. ุงุนุฑู ุงู„ู…ุฒูŠุฏ ุนู† ุณูŠุงุณุฉ ุงู„ุฎุตูˆุตูŠุฉ