Question

★ CONTENT QUALITY SUPPORT REQUIRED ★

★ STANDARD QUESTION 5 ★

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Question is -

Solve for the real " x "

$\sqrt{x + 3 - 4 \sqrt{x - 1} } + \sqrt{x + 8 - 6 \sqrt{x - 1} } = 1$

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Answer 1

x € [5,10] this might help you

Answer 2

Given,

$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\\\\Let,\\\\\sqrt{x-1}=y\implies\;x=y^{2}+1\\ \\\text{Putting these values,}\\\\\sqrt{y^{2}+1+3-4y}+\sqrt{y^{2}+1+8-6y}=1\\\\\sqrt{y^{2}+4-4y}+\sqrt{y^{2}+9-6y}=1\\\\\sqrt{y^{2}+2^{2} -2\times2\times y}+\sqrt{y^{2}+3^{2}-2\times3\times y}=1\\\\\sqrt{(y-2)^{2}}+\sqrt{(y-3)^{2}}=1\\\\\frac{+}{}(y-2)+\frac{+}{}(y-3)=1$

Taking  +ve and  +ve

$y-2+y-3=1\\\\2y-5=1\\\\2y=6\\\\y=3$

Taking +ve and -ve

$y-2-y+3=1\\\\2y+1=1\\\\y=0$

Taking  -ve and  +ve

$-y+2+y-3=1\\\textbf{No solution possible}$

Taking  -ve and  -ve

$-y+2-y+3=1\\\\-2y+5=1\\\\2y=4\\\\y=2$

Now , We have  y = 3 , 0 , 2, Putting  this values  in x = y²+1

(i) when y = 3 ,

x = 3² + 1

x = 10

(ii)  when y = 0

x  = 0 +1

x = 1 , but this  will not  satisfy  equation,

x≠1

(iii) when y = 2

x  = 2² + 1

x = 5

hence,

x = 5 or 10