How to solve 7(x-4)2-3(x+5)2=4(x+1)(x-1)-2 ?

Welcome to my article 7(x-4)2-3(x+5)2=4(x+1)(x-1)-2 This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question ,7(x-4)2-3(x+5)2=4(x+1)(x-1)-2 read and understand it carefully till the end.

Let us know how to solve this question7(x-4)2-3(x+5)2=4(x+1)(x-1)-2

First write the question on the page of the notebook

7(x-4)2-3(x+5)2=4(x+1)(x-1)-2

\displaystyle (7\text{x}x-7\text{x}4)2-(3\text{x}x+3\text{x5)2}=4\left( {x+1} \right)\left( {x-1} \right)-2

\displaystyle (7x-28)2-(3x+1\text{5)2}=4\left( {x+1} \right)\left( {x-1} \right)-2

\displaystyle (7x\text{x}2-28\text{x2})-(6x+15\text{x2)}=4\left( {x+1} \right)\left( {x-1} \right)-2

\displaystyle (14x-56)-(6x+30\text{)}=4\left( {x+1} \right)\left( {x-1} \right)-2

\displaystyle 14x-56-6x-30=4\left( {x+1} \right)\left( {x-1} \right)-2

\displaystyle 14x-6x-30-56=4\left( {x+1} \right)\left( {x-1} \right)-2

\displaystyle 8x-86=4\left( {x+1} \right)\left( {x-1} \right)-2

we know that,

\displaystyle {{x}^{2}}-{{1}^{2}}=(x+1)(x-1)

than,

\displaystyle 8x-86=4({{x}^{2}}-{{1}^{2}})-2

\displaystyle 8x-86=4{{x}^{2}}-4-2

\displaystyle 8x-86=4{{x}^{2}}-6

\displaystyle 8x-4{{x}^{2}}+6-86=0

\displaystyle 8x-4{{x}^{2}}-80=0

\displaystyle -4{{x}^{2}}+8x-80=0

\displaystyle -(4{{x}^{2}}-8x+80)=0

\displaystyle 4{{x}^{2}}-8x+80=0

\displaystyle 4({{x}^{2}}-2x+20)=0

\displaystyle {{x}^{2}}-2x+20=0

Use this formula to solve this question,

\displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}

\displaystyle \begin{array}{l}a=1,\b=-2\c=20\end{array}

\displaystyle x=\frac{{-(-2)\pm \sqrt{{{{{(-2)}}^{2}}-4.1.20}}}}{{2.1}}

\displaystyle x=\frac{{2\pm \sqrt{{4-80}}}}{{2.1}}

\displaystyle x=\frac{{2\pm \sqrt{{-76}}}}{{2.1}}

\displaystyle x=\frac{{2+\sqrt{{-76}}}}{{2.1}}Answer

\displaystyle x=\frac{{2-\sqrt{{-76}}}}{{2.1}}Answer

or,

\displaystyle x=\frac{{2+\sqrt{{2.2.-19}}}}{{2.1}}

\displaystyle x=\frac{{2+2\sqrt{{-19}}}}{2}

\displaystyle x=\frac{{2(1+\sqrt{{-19)}}}}{2}

\displaystyle x=(1+\sqrt{{-19)}}Answer

\displaystyle x=(1-\sqrt{{-19)}}Answer

7(x-4)2-3(x+5)2=4(x+1)(x-1)-2This article has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.

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To see other examples like this see

2/x+6+7/x+9?x=5/x-2/x

To solve such questions, first of all put the variable amount terms together as follows and keep the constant amount terms together.

\displaystyle \frac{2}{x}+\frac{7}{x}+\frac{9}{x}-\frac{5}{x}+\frac{2}{x}=-6<em>x</em>2​+<em>x</em>7​+<em>x</em>9​−<em>x</em>5​+<em>x</em>2​=−6

Since the denominators are equal, add the numerators together in the following way.

\displaystyle \frac{{2+7+9-5+2}}{x}=-6x2+7+9−5+2​=−6

\displaystyle \frac{{20-5}}{x}=-6x20−5​=−6

Change sides.

\displaystyle \frac{{15}}{x}+6=0x15​+6=0

\displaystyle \frac{{15+6x}}{x}=0x15+6x​=0

Since multiplying 0 by any number gives 0. so 0xX = 0

\displaystyle 15+6x=015+6x=0

\displaystyle 6x=-156x=−15

\displaystyle x=-\frac{{15}}{6}x=−615​

-15 /6 can also be written as 5×3 /2×3. If there is a similar number from the bottom up and down, then remove it.

\displaystyle x=\frac{{-5}}{2}Answerx=2−5​Answer

How to solve 1/x+1+2/x+2=4/x+4 ? This article has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.

Note: If you have any such question, then definitely send it by writing in our comment box to get the answer.
Your question will be answered from our side.

Thank you once again from our side for reading or understanding this article completely.