Question

Solve the following equation a plus b the whole square x square - 4 ab x minus a minus b the whole square is equals to zero​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given:

An equation, "a plus b the whole square x square - 4 ab x minus a minus b the whole square equals to zero​".

To Find:

The solution of the equation.

Solution:

The equation is given in the form of a sentence, write it in an equation containing signs and terms.

$(a+b)^2{x^2} -4abx-(a-b)^2=0$

In the above equation, $-4abx$ can be written as $-4abx=-(a+b)^2x+(a-b)^2x$.

Use the result $-4abx=-(a+b)^2x+(a-b)^2x$ and re-write the above equation.

$(a+b)^2{x^2}-(a+b)^2x+(a-b)^2x-(a-b)^2=0$

Take out common terms from the expression.

$(a+b)^2x(x-1)+(a-b)^2(x-1)=0$

Take out $(x+1)$ common from the whole expression.

$(x-1)((a+b)^2x+(a-b)^2)=0$

Equate $(x-1)$ to 0 and find the roots.

$(x-1)=0\\x=1$

Equate $(a+b)^2x+(a-b)^2$ to zero and find roots.

$(a+b)^2x+(a-b)^2=0\\\\x=\frac{-(a-b)^2}{(a+b)^2}$

The obtained roots will satisfy the given equation.

Check by substituting $x=1$ into the equation$(a+b)^2{x^2} -4abx-(a-b)^2=0$.

$(a+b)^2{1^2} -4ab\cdot 1-(a-b)^2=0\\(a+b)^2-4ab-(a-b)^2=0\\a^2+2ab+b^2-4ab-(a^2+2ab+b^2)=0\\a^2+2ab+b^2-4ab-a^2-2ab-b^2=0\\0=0\\{\rm LHS}={\rm RHS}$

$x=1$ is the root of the equation because it is satisfying the equation.

Check by substituting $x=\frac{-(a-b)^2}{(a+b)^2}$ into the equation $(a+b)^2{x^2} -4abx-(a-b)^2=0$.

$(a+b)^2{(-\frac{(a-b)^2}{(a+b)^2})^2} -4ab{(-\frac{(a-b)^2}{(a+b)^2})}-(a-b)^2=0\\(a-b)^4+4ab(a-b)^2-(a-b)^2(a+b)^2=0\\(a-b)^2(a^2+b^2-2ab+4ab-(a^2+b^2+2ab))=0\\(a-b)^2(a^2+b^2-2ab+4ab-a^2-b^2-2ab)=0\\0=0\\{\rm LHS}={\rm RHS}$

$x=\frac{-(a-b)^2}{(a+b)^2}$ is also the root of the equation because it is satisfying the equation.

Thus, the solution of the given equation is $x=1$ and$x=\frac{-(a-b)^2}{(a+b)^2}$.