Question 18

If$$\ a^{2}+b^{2}+c^{2}$$= 2(a-b-c)-3, then the value of 4a - 3b + 5c is

Solution

$$\ a^{2}+b^{2}+c^{2}$$= 2(a-b-c)-3
We can write the above equation as

$$ a^{2}-2a+1+b^{2}+2b+1+c^{2}+2c+1$$=0
$$\Rightarrow (a-1)^{2}+(b+1)^{2}+(c+1)^{2}$$=0

$$(a-1)^{2}$$=0$$\Rightarrow$$a=1
$$(b+1)^{2}$$=0$$\Rightarrow$$ b=-1
$$(c+1)^{2}$$=0$$\Rightarrow$$ c=-1

4a-3b+5c= 4(1)-3(-1)+5(-1)=4+3-5=2

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