Question
Determine. if 3 is a root of the equation given below:
$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
✓
Answer
$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
If x = 3, then
L.H.S. $\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}$
$=\sqrt{(3)^2-4\times3+3}+\sqrt{(3)^2-9}$
$=\sqrt{9-12+3}+\sqrt{9-9}$
$=\sqrt{0}+\sqrt{0}$
$=0$
R.H.S. $=\sqrt{4\text{x}^2-14\text{x}+16}$
$=\sqrt{4(3)^2-14\times3+16}$
$=\sqrt{4\times9-42+16}$
$=\sqrt{36+16-42}$
$=\sqrt{52-42}$
$=\sqrt{10}$
$\because$ L.H.S. $\neq$ R.H.S.
If x = 3, then
L.H.S. $\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}$
$=\sqrt{(3)^2-4\times3+3}+\sqrt{(3)^2-9}$
$=\sqrt{9-12+3}+\sqrt{9-9}$
$=\sqrt{0}+\sqrt{0}$
$=0$
R.H.S. $=\sqrt{4\text{x}^2-14\text{x}+16}$
$=\sqrt{4(3)^2-14\times3+16}$
$=\sqrt{4\times9-42+16}$
$=\sqrt{36+16-42}$
$=\sqrt{52-42}$
$=\sqrt{10}$
$\because$ L.H.S. $\neq$ R.H.S.
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