Question

$(2x + 3) {}^{2} =81$
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Answer 1

Given:

• $\sf{(2x+3)^{2}=81}$

To Find:

• The value of x .

Concept Used:

• ☞First- LHS is expressed in the form of square of a expression .So ,we will try to obtain the RHS also in the terms of square of a number so that square root can be eliminated both sides.
• ☞Second- We will square root both sides ,then ,after simplification of square root we will proceed.

Answer:

Some points to be noted:

1. If we used first method ,then we will get only one value of x .
2. If we use second method we will get 2 values of x in which one of its value will be equal to first value.

So ,we will proceed by second method.

☞Taking the given equation;

$\sf{\implies (2x+3)^{2}=81}$

Putting square root both sides:

$\sf{\implies\sqrt{(2x+3)^{2}}=\sqrt{81}}$

$\sf{\implies (2x+3)=\sqrt{81}}$

$\sf{\implies (2x+3)=\pm 9}$

☞Taking (+9) = 2x+3 .

$\sf{\implies 2x+3=9}$

$\sf{\implies 2x=9-3}$

$\sf{\implies 2x =6}$

$\sf{\implies x=\cancel{\dfrac{6}{2}}}$

${\underline{\boxed{\red{\sf{\leadsto x=3}}}}}$

☞ Taking (-9)=2x +3.

$\sf{\implies 2x+3=(-9)}$

$\sf{\implies 2x=-(9+3)}$

$\sf{\implies 2x =(-12)}$

$\sf{\implies x=\cancel{\dfrac{-12}{2}}}$

${\underline{\boxed{\red{\sf{\leadsto x=(-6)}}}}}$

Hence we got two values of x which are (-6) & 3.