Question
In the following, determine whether the given values are solution of the given equation or not:
$2x^2 - x + 9 = x^2 + 4x + 3, x = 2, x = 3$
$2x^2 - x + 9 = x^2 + 4x + 3, x = 2, x = 3$
✓
Answer
$2 x^2-x+9=x^2+4 x+3, x=2, x=3 \text { When, } x=2$
Substituting $x=2$
$\text { L.H.S. }=2 x^2-x+9=2(2)^2-2+9$
$=8-2+9=15$
$\text { R.H.S. }=x^2+4 x+3=(2)^2+4 \times 2+3$
$=4+8+3=15$
$\because \text { L.H.S. }=\text { R.H.S. }$
$\therefore x =2$ is the solution
When, $x=3$
$\text { L.H.S. }=2 x^2-x+9$
$=2(3)^2-3+9$
$=18-3+9=24$
$\text { R.H.S. }=x^2+4 x+3$
$=(3)^2+4 \times 3+3$
$=9+12+3$
$=24$
$\because \text { L.H.S. }=\text { R.H.S. }$
$\therefore x=3$ is ths solution.
Substituting $x=2$
$\text { L.H.S. }=2 x^2-x+9=2(2)^2-2+9$
$=8-2+9=15$
$\text { R.H.S. }=x^2+4 x+3=(2)^2+4 \times 2+3$
$=4+8+3=15$
$\because \text { L.H.S. }=\text { R.H.S. }$
$\therefore x =2$ is the solution
When, $x=3$
$\text { L.H.S. }=2 x^2-x+9$
$=2(3)^2-3+9$
$=18-3+9=24$
$\text { R.H.S. }=x^2+4 x+3$
$=(3)^2+4 \times 3+3$
$=9+12+3$
$=24$
$\because \text { L.H.S. }=\text { R.H.S. }$
$\therefore x=3$ is ths solution.
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