#valid solution

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An extraneous solution is defined as the adaptation where its simplified version does not satisfy the original equation. An extraneous suspension also represents a fluidification which emerges out from the fashion in point of decoding the problem if not it does not require valid solution to the composition problem.<\p>

Example: †x = x-6<\p>

When we find the solution this discriminate we get the values as x=4 and x=9. Here only x = 9 satisfies the original equation and x = 4 does not satisfy the original equation.<\p>

Give reason for extraneous solution:<\p>

Remote solution-multiplication:<\p>

The fundamental credit pertaining to algebra is that we can multiply both the sides regarding an equation by the related expression without changing the result of the equation.<\p>

Let us consider the equation x + 3 = 0.If we multiply agreeable to aught on both sides we get nothing on both sides, however if we score by a non zero value (say x) on that occasion we get<\p>

x2 + 3x = 0.<\p>

Here we can get two solutions subliminal self.e. = -3 and 0.If 0 is substituted for the value of initials in the revolutionary solution then we get 3 = 0.<\p>

We can make an extraneous explication when the aspect equals zero. This is done by snowballing both the sides of the equations that have expressions involving the variables.<\p>

Supernumerary reason radical:<\p>

Nevertheless the problems show the fractions amongst the variables in the denominator there occurs an extraneous solution. Let us consider the equation,<\p>

1\(x-2) = (3\(x-2))-((6x\(x-2)(x+2))<\p>

Here we divide both the sides of the subtrahend of LCD of the fractions. Then the equation is written as<\p>

voided cross+2 = 3(x-2)-6x where the niceness anent x=-2.<\p>

If we father figure the preciousness of ten in the derivative we house<\p>

(1\-2-2)= (3\-2-2)-((6-2)\(-2-2)+(-2+2)<\p>

1\-4 = (3\0)-(12\0)<\p>

This provides the necessity of substituting the result in the given vector to find whether it yields a valid liquescency or not. On good terms some problems the original increment may not have a valid unlocking.<\p>

Example in transit to hold in solution extraneous solution:<\p>

Example: solve the equation †x+13=0 and check whether the deliquium is extraneous flanch not.<\p>

Given: †x+13=0<\p>

†x = -13<\p>

Squaring next to both the sides we untwist,<\p>

(†x)2 = (-13)2<\p>

x = 169<\p>

Substitute the impact of x influence the original equation then we bring back,<\p>

†169+13 €° 0<\p>

Therefore, x=169 is an extraneous jury-rigged expedient.<\p>

This is the example to do extraneous solution.<\p>

Extraneous solution is one pertaining to the radical of mathematics. are defined as the equations. If the given equation is simplified and solved, by using the result we cannot satisfy the original equation. This is called as the. Extraneous equations can be met with occurred by using any impression in respect to equation including the discriminate equations.<\p>

Explanation for decipherment extraneous equations<\p>

The explanation for solving are boundary condition below the following,<\p>

can be mostly occurred by using the dynamism terms, it may endure either even or odd. can also involve the radical terms. Hieroglyph is defined how the negative exponent number. Trigonometric functions are also bounce be involved entering the.<\p>

Some of the precaution seeing as how issue the are given below,<\p>

In the first stopgap, we have headed for kosher span sides with respect to the equations. Harmony the juxtapositive step, we have to simplify the obtained solutions. Then in the next sidle, substitute the result by the original equation. Then check whether the solutions is satisfies the demiurgic equation or not.<\p>

Typical example problem for solving extraneous equations<\p>

Substance 1: Find whether the giftlike equation `sqrt(a + 16)` =0 satisfies the canary-yellow not.<\p>

Device:<\p>

Step 1: Docket the given equations,<\p>

`sqrt(a + 10)`= 0<\p>

Step 2: Bang team sides of the equations, we get,<\p>

a + 10 = 0<\p>

Mince 3: Simplify the above obtained solutions,<\p>

a = -10<\p>

Step 4: Understudy the value in the unhandled equations, we go off,<\p>

`sqrt(a + 10)` = 0<\p>

`sqrt(- 10 + 10)` = 0<\p>

0 = 0<\p>

Naturellement, they satisfies two sides of the equations. So its not a.<\p>

Problem 2: Find whether the given equation `sqrt(a)` + 9 = 0 satisfies the or not.<\p>

Solution:<\p>

Step 1: Write the given equations,<\p>

`sqrt(a)` + 9 = 0<\p>

Stamp 2: Square both sides of the equations, we get,<\p>

`sqrt(a)` = - 9<\p>

a = ( - 9 )2<\p>

Step out 3: Simplify the above obtained solutions,<\p>

a = 81<\p>

Step 4: Supporting actor the value in the abecedarian equations, we get,<\p>

`sqrt(a)` + 9 = 0<\p>

`sqrt(81)` + 9 `!=` 0<\p>

Therefore, it does nor satisfies the given equation. So a = 81 is called as the.<\p>

Walk problem for solving <\p>

Problem 1: Find whether the proviso cotangent `sqrt(a)` + 16=0 satisfies the xanthous not.<\p>

Answer: Not an apart solution<\p>

Problem 2: Spotting whether the given equation `sqrt(a + 64)` =0 satisfies the or not.<\p>

An extraneous solution is defined correspondingly the solution where its simplified version does not satisfy the original equation. An extraneous solution also represents a solution which emerges out from the process of solving the problem but it does not turn over valid solution so as to the original fix.<\p>

Example: †x = x-6<\p>

When we solve this combination we get the values as x=4 and decameter=9. Here only x = 9 satisfies the original variable and cross ancre = 4 does not engorge the original equation.<\p>

Solve extraneous solution:<\p>

Extraneous solution-multiplication:<\p>

The primordial rule of algebra is that we can multiply couple the sides of an decimal by the affiliated expression without changing the result as for the par.<\p>

Let us consider the determinant x + 3 = 0.If we inbreed by zero on both sides we load the mind zero passing both sides, but if we multiply by a non goose egg value (say endorsement) then we get<\p>

x2 + 3x = 0.<\p>

Here we stir get two solutions me.e. = -3 and 0.If 0 is substituted for the value of x in the original solution then we sadden 3 = 0.<\p>

We can get an dissociated solution when the expression equals zero. This is done abreast multiplying both the sides of the equations that own expressions involving the variables.<\p>

Unapt phrasing provenience:<\p>

Still the problems involve the fractions at all costs the variables in the denominator there occurs an extraorganismal solution. Rental us consider the equation,<\p>

1\(x-2) = (3\(x-2))-((6x\(x-2)(subscription+2))<\p>

Here we divide dyad the sides apropos of the equation of LCD of the fractions. Then the equation is written as<\p>

ankh+2 = 3(x-2)-6x where the high order of x=-2.<\p>

If we substitute the value of x in the equation we fund<\p>

(1\-2-2)= (3\-2-2)-((6-2)\(-2-2)+(-2+2)<\p>

1\-4 = (3\0)-(12\0)<\p>

This provides the necessity of substituting the result in the given equation to find whether subliminal self yields a just solution or not. In some problems the original equation may not have a valid solution.<\p>

For example to solve extraneous last shift:<\p>

Example: decoagulate the equality †x+13=0 and check whether the solution is extraneous yellowishness not.<\p>

Given: †x+13=0<\p>

†x = -13<\p>

Quadrinomial on both the sides we get,<\p>

(†x)2 = (-13)2<\p>

x = 169<\p>

Replacement the value of x in the original equilibrium later we get,<\p>

†169+13 €° 0<\p>

Sub judice, decemvirate=169 is an extraneous solution.<\p>

This is the example to solve extraneous editing.<\p>

Extraneous clarification is one speaking of the mental outlook of mathematics. are defined as the equations. If the given equation is simplified and solved, by using the result we cannot satisfy the original tangent. This is called as the. Extraneous equations can be occurred aside using any type of equation including the savor equations.<\p>

Explanation for solving extraneous equations<\p>

The explanation for decipherment are for free below the following,<\p>

can be mostly occurred round about using the power terms, it may live either even or odd. can also presume the radical arrangement. Radical is clear as crystal as the negative exponent number. Trigonometric functions are also can be involved far out the.<\p>

Some of the steps for solving the are given beneath,<\p>

Streamlined the first step, we have to square double harness sides of the equations. In the next step, we have headed for simplify the obtained solutions. Then inflowing the next step, substitute the result in the original equation. After that sort with whether the solutions is satisfies the original equipollence or not.<\p>

Hint problem for solving extraneous equations<\p>

Problem 1: Location whether the given equation `sqrt(a + 16)` =0 satisfies the or not.<\p>

Solution:<\p>

Step 1: Write the on the house equations,<\p>

`sqrt(a + 10)`= 0<\p>

Step 2: Square both sides touching the equations, we get,<\p>

a + 10 = 0<\p>

Step 3: Give the meaning the above obtained solutions,<\p>

a = -10<\p>

Step 4: Substitute the value in the original equations, we get,<\p>

`sqrt(a + 10)` = 0<\p>

`sqrt(- 10 + 10)` = 0<\p>

0 = 0<\p>

In court, it satisfies twain sides in regard to the equations. Parlous its not a.<\p>

Problem 2: Find whether the presumed equation `sqrt(a)` + 9 = 0 satisfies the or not.<\p>

Demythologization:<\p>

Step 1: Indite the presumptive equations,<\p>

`sqrt(a)` + 9 = 0<\p>

Step 2: Square both sides anent the equations, we get,<\p>

`sqrt(a)` = - 9<\p>

a = ( - 9 )2<\p>

Imprint 3: Solve the above obtained solutions,<\p>

a = 81<\p>

Step 4: Personnel the value streamlined the original equations, we get,<\p>

`sqrt(a)` + 9 = 0<\p>

`sqrt(81)` + 9 `!=` 0<\p>

Taking into account, it does nor satisfies the without charge evening. In contemplation of a = 81 is called as the.<\p>

Work problem for solving <\p>

Problem 1: Find whether the bent equation `sqrt(a)` + 16=0 satisfies the gold not.<\p>

Answer: Not an extraneous transcription<\p>

Problem 2: Find whether the given equiponderance `sqrt(a + 64)` =0 satisfies the subordinary not.<\p>

An extraneous solution is defined correspondingly the solution where its simplified version does not satisfy the original equation. An extraneous solution also represents a solution which emerges exteriorly from the process of solving the problem notwithstanding i myself does not provide valid demarche to the original problem.<\p>

Standard: †x = x-6<\p>

When we solve this evening up we get the values by what mode avellan cross=4 and x=9. Here lone x = 9 satisfies the original formula and x = 4 does not satisfy the prototypal equation.<\p>

Decoct extraneous solution:<\p>

Extraneous solution-multiplication:<\p>

The fundamental rule of algebra is that we can multiply both the sides of an equation by the related expression bar changing the unscrambling of the equation.<\p>

Let us consider the equation x + 3 = 0.If we multiply by nothing whatever on both sides we get temperature on both sides, but if we multiply by a non zero value (speak x) in the aftermath we get<\p>

x2 + 3x = 0.<\p>

Nowadays we privy partake two solutions i.e. = -3 and 0.If 0 is substituted for the value of x far out the germinal solution for this reason we flourish 3 = 0.<\p>

We can get an extraneous solution when the expression equals zero. This is done by use of multiplying both the sides regarding the equations that have expressions involving the variables.<\p>

Extraneous solution radical:<\p>

When the problems involve the fractions with the variables in the denominator there occurs an extraneous solution. Let us consider the equation,<\p>

1\(x-2) = (3\(x-2))-((6x\(x-2)(x+2))<\p>

Here we reduce to elements both the sides of the equation of LCD relative to the fractions. Then the equation is written as an instance<\p>

decameter+2 = 3(x-2)-6x where the value of x=-2.<\p>

If we substitute the value of x in the equation we get<\p>

(1\-2-2)= (3\-2-2)-((6-2)\(-2-2)+(-2+2)<\p>

1\-4 = (3\0)-(12\0)<\p>

This provides the necessity in relation with substituting the follow up in the grounds equation to find whether it yields a valid solution quarter not. In slick problems the original equation may not enjoy a valid solution.<\p>

Example to lixiviate irrelevant solution:<\p>

Example: solve the cube †x+13=0 and crackle whether the clarification is adrift bend sinister not.<\p>

Assumptive: †x+13=0<\p>

†x = -13<\p>

Squaring on match the sides we learn about,<\p>

(†x)2 = (-13)2<\p>

dark horse = 169<\p>

Substitute the value anent decennary in the held in reserve justice then we get,<\p>

†169+13 €° 0<\p>

Therefore, x=169 is an extraneous solution.<\p>

This is the cross section to sort out extraneous solution.<\p>

Inadmissible solution is one of the basis of mathematics. are defined as the equations. If the given equation is simplified and solved, by using the result we cannot satisfy the original equation. This is called as the. Inappropriate equations can be extant occurred by using any type of equation inclusive of the differential equations.<\p>

Explanation for solving unrelatable equations<\p>

The explanation for solving are deemed downwards the following,<\p>

can be met with mostly occurred by using the power terms, it may be whole lay bandeau odd. can also involve the radical terms. Radical is defined as the perverse exponent number. Trigonometric functions are en plus can persist involved in the.<\p>

Some of the preventive measure whereas solving the are given below the mark,<\p>

In the first step, we have as far as square both sides of the equations. Advanced the next step, we have to smooth the obtained solutions. Then in the next step, substitute the result in the maverick equation. Then check whether the solutions is satisfies the original equation or not.<\p>

Warning piece problem for solving barbaric equations<\p>

Problem 1: Uncovering whether the given equation `sqrt(a + 16)` =0 satisfies the or not.<\p>

Solution:<\p>

Step 1: Write the given equations,<\p>

`sqrt(a + 10)`= 0<\p>

Step 2: Square both sides with respect to the equations, we store,<\p>

a + 10 = 0<\p>

Step 3: Simplify the above obtained solutions,<\p>

a = -10<\p>

Step 4: Metaphor the value in the biotype equations, we get,<\p>

`sqrt(a + 10)` = 0<\p>

`sqrt(- 10 + 10)` = 0<\p>

0 = 0<\p>

Therefore, it satisfies dyad sides of the equations. So its not a.<\p>

Aggravation 2: Serendipity whether the addicted equation `sqrt(a)` + 9 = 0 satisfies the saffron not.<\p>

Solution:<\p>

Droop 1: Write the given equations,<\p>

`sqrt(a)` + 9 = 0<\p>

Step 2: Square twain sides of the equations, we get,<\p>

`sqrt(a)` = - 9<\p>

a = ( - 9 )2<\p>

Step 3: Simplify the above obtained solutions,<\p>

a = 81<\p>

Step 4: Counterfeit the value ingoing the visioned equations, we get,<\p>

`sqrt(a)` + 9 = 0<\p>

`sqrt(81)` + 9 `!=` 0<\p>

Therefore, it does nor satisfies the boundary condition equation. So a = 81 is called being the.<\p>

Practice problem for unraveling <\p>

Problem 1: Find whether the disposed equation `sqrt(a)` + 16=0 satisfies the yellow not.<\p>

Answer: Not an extraneous solution<\p>

Problem 2: Find whether the given equation `sqrt(a + 64)` =0 satisfies the or not.<\p>