In the following, determine whether the given values are solutions of the given equation or not:
$2x^2\ β\ x\ +\ 9\ =\ x^2\ +\ 4x\ +\ 3,\ x\ =\ 2,\ x\ =\ 3$
Given:
The given equation is $2x^2\ –\ x\ +\ 9\ =\ x^2\ +\ 4x\ +\ 3$
To do:
We have to determine whether $x=2, x=3$ are solutions of the given equation.
Solution:
If the given values are the solutions of the given equation then they should satisfy the given equation.
Therefore,
For $x=2$,
LHS$=2x^2-x+9$
$=2(2)^2-2+9$
$=8-2+9$
$=15$
RHS$=x^2+4x+3$
$=(2)^2+4(2)+3$
$=4+8+3$
$=15$
LHS$=$RHS
Hence, $x=2$ is a solution of the given equation.
For $x=3$,
LHS$=2x^2-x+9$
$=2(3)^2-3+9$
$=18-3+9$
$=24$
RHS$=x^2+4x+3$
$=(3)^2+4(3)+3$
$=9+12+3$
$=24$
LHS$=$RHS
Hence, $x=3$ is a solution of the given equation.β
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