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Solve the following equation and also check your result:
$[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$
Given:
The given equation is $[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$
To do:
We have to solve the given equation and check the result.
Solution:
To check the result we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.
The given equation is $[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$
$[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$
$[3x+8]^2+[x-2]^2=10x^2 + 92$
$(3x)^2+2(3x)(8)+8^2+x^2-2(x)(2)+2^2=10x^2+92$ [Since $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$]
$9x^2+48x+64+x^2-4x+4=10x^2+92$
$10x^2+44x+68=10x^2+92$
On rearranging, we get,
$10x^2-10x^2+44x=92-68$
$44x=24$
$x=\frac{24}{44}$
$x=\frac{6}{11}$
Verification:
LHS $=[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2$
$=[(2(\frac{6}{11})+3)+(\frac{6}{11}+5)]^2+[(2(\frac{6}{11})+3)-(\frac{6}{11}+5)]^2$
$=[\frac{12}{11}+3)+(\frac{6}{11}+5)]^2+[(\frac{12}{11})+3)-(\frac{6}{11}+5)]^2$
$=[\frac{12+6}{11}+8]^2+[\frac{12-6}{11}-2]^2$
$=[\frac{18}{11}+8]^2+[\frac{6}{11}-2]^2$
$=[\frac{18+11\times8}{11}]^2+[\frac{6-2\times11}{11}]^2$
$=[\frac{18+88}{11}]^2+[\frac{6-22}{11}]^2$
$=[\frac{106}{11}]^2+[\frac{-16}{11}]^2$
$=\frac{11236}{121}+\frac{256}{121}$
$=\frac{11236+256}{121}$
$=\frac{11492}{121}$
RHS $=10x^2 + 92$
$=10(\frac{6}{11})^2 + 92$
$=10(\frac{36}{121})+92$
$=\frac{360}{121}+92$
$=\frac{360+121\times92}{121}$
$=\frac{360+11132}{121}$
$=\frac{11492}{121}$
LHS $=$ RHS
Hence verified.
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