Question

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Answer 1

Answer:

Step-by-step explanation:

Question is

What Should Be Added to $x^{4}$+$2x^{2}$-7  to obtain $4x^{4}$-$3x^{3} +x^{2}$+2

I'am Just Explaining to you By the Example in Numbreic Value

So Let thinks It May Be

What Should Be Added to 6   to obtain 8

So WE will Simply Do this,

8 - 6=2

So If We Add 2 To 6 So we will Get 8

In This Way, We will solve Your First Question

$4x^{4} -3x^{3}+ x^{2} +2$ -[ $x^{4}+2x^{2}-7$]

$4x^{4} -3x^{3}+ x^{2} +2$ $-x^{4}-2x^{2}+7$   [After Opening Bracket Sings Were Changed]

$4x^{4} -x^{4}$$-3x^{3}$$+ x^{2}-2x^{2}+2+7$

$3x^{4}-3x^{3}-x^{2}+9$

So

 $3x^{4}-3x^{3}-x^{2}+9$ Should Be Added to $x^{4}+2x^{2}$-7  to obtain $4x^{4}-3x^{3} +x^{2}$+2

Answer 2

FINAL ANSWER:

4.     $3x^{4}-3x^{3}-x^{2}+9$

5.     $-5xy^{2}+4x^{2}y-3x^{2}y^{2}-7xy$

THINGS TO CONSIDER:-

4.     a+b=c

5.     a-b=c

GIVEN:

4.   a=$x^{4}+2x^{2}-7$

     c=$4x^{4}-3x^{3}+x^{2}+2$

5.   a=$4x^{2}y^{2}+3xy+3xy^{2}$

     c=$8xy^{2}-4x^{2}y+7x^{2}y^{2}+10xy$

TO FIND:

4.    b=?

5.    b=?

SOLUTION:

4.   a+b=c

$x^{4}+2x^{2}-7+b=4x^{2}y^{2}+3xy+3xy^{2}$

b=$4x^{4}-3x^{3}+x^{2}+2-(x^{4}+2x^{2}-7)$

b=$4x^{4}-3x^{3}+x^{2}+2-x^{4}-2x^{2}+7)$

b=$3x^{4}-3x^{3}-x^{2}+9$

5.   a-b=c

$4x^{2}y^{2}+3xy+3xy^{2}-(b)=8xy^{2}-4x^{2}y+7x^{2}y^{2}+10xy$

$-(b)=8xy^{2}-4x^{2}y+7x^{2}y^{2}+10xy-(4x^{2}y^{2}+3xy+3xy^{2})$

$-(b)=8xy^{2}-4x^{2}y+7x^{2}y^{2}+10xy-4x^{2}y^{2}-3xy-3xy^{2}$

$-(b)=5xy^{2}-4x^{2}y+3x^{2}y^{2}+7xy$

[↑↑↑↑↑ when i open b bracket so minus goes to RHS and than whole side signs get changed]

$b=-5xy^{2}+4x^{2}y-3x^{2}y^{2}-7xy$

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