Question

$\huge{\bf{\underbrace{\orange{Question}}}}$
solve for x -
$\sqrt{2x + 7} = x + 2$
​

Answer 1

Working out:

A equation is given in the question having the variable of x in both sides. We have to solve for x?

GiveN :

• $\sf{ \sqrt{2x + 7} = x + 2}$

So, let's start solving the above:

$\sf{ \longrightarrow{ \sqrt{2x + 7} = x + 2}}$

Squaring both sides,

$\sf{ \longrightarrow{( \sqrt{2x + 7}) {}^{2} = (x + 2) {}^{2} }}$

By using the identity of (a + b)² = a² + 2ab + b² in RHS,

$\sf{ \longrightarrow{2x + 7 = {x}^{2} + 4x + 4}}$

Now shifting all the terms to the LHS,

$\sf{ \longrightarrow{2x + 7 - {x}^{2} - 4x - 4 = 0}}$

$\sf{ \longrightarrow{ - {x}^{2} - 2x + 3= 0}}$

Multiplying with a negative sign both sides,

$\sf{ \longrightarrow{ {x}^{2} + 2x - 3 = 0}}$

Factorising by using middle term factorisation,

$\sf{ \longrightarrow{ {x}^{2} + 3x - x - 3 = 0}}$

$\sf{ \longrightarrow{x(x + 3) - 1(x + 3) = 0}}$

$\sf{ \longrightarrow{(x - 1)(x + 3) = 0}}$

So, x can be 1 or -3

Both the values are possible since they are satisfying the above equation. Hence, the answer is:

$\huge{ \boxed{ \sf{ \red{x = 1 \: or \: - 3}}}}$

And we are done !!

Answer 2

$\sf{\blue{\underline{\underline{\pink{GIVEN:-}}}}}$

• $\bf{\sqrt{2x + 7} = x + 2}$

$\sf{\blue{\underline{\underline{\pink{TO\: FIND:-}}}}}$

• The value of x .

$\sf{\blue{\underline{\underline{\pink{SOLUTION:-}}}}}$

$\green\checkmark\:\bf{\sqrt{2x + 7} = x + 2}$

$\rm{\implies\:2x\:+\:7\:=\:(x\:+\:2)^2}$

$\rm{\implies\:2x\:+\:7\:=\:x^2\:+\:2\times{x}\times{2}\:+\:2^2\:}$

$\rm{\implies\:x^2\:+\:4x\:+\:4\:=\:2x\:+\:7\:}$

$\rm{\implies\:x^2\:+\:4x\:-\:2x\:+\:4\:-\:7\:=\:0\:}$

$\rm\red{\implies\:x^2\:+\:2x\:-\:3\:=\:0\:}$

$\rm{\implies\:x^2\:+\:3x\:-\:x\:-\:3\:=\:0\:}$

$\rm{\implies\:x\:(x\:+\:3)\:-1\:(x\:+\:3)\:=\:0}$

$\rm{\implies\:(x\:+\:3)\:(x\:-\:1)\:=\:0}$

$\rm{\implies\:x\:+\:3\:=\:0\:\:\:\:or\:\:\implies\:x\:-\:1\:=\:0\:}$

$\rm\pink{\implies\:x\:=\:-3\:\:\:\:or\:\:\implies\:x\:=\:1\:}$

$\red\therefore\:\bf{\gray{\underline{\red{\boxed{\green{The\:values\:of\:x\:are\:-3\:and\:1\:}}}}}}$