Question

If \[100\dfrac{\sqrt{25}}{\sqrt{25}+x}=50,\] then the value of x is:
(a) 25
(b) \[\dfrac{1}{\sqrt{25}}\]
(c) \[\sqrt{25}\]

Answer 1

Hint: To solve the question given above, we will first divide the whole equation by 100. Then we will rationalize the left-hand side of the equation. After rationalizing, we will get a quadratic in x. We will solve it with the help of the factorization method and find the value of x.

Complete step-by-step answer:
In the first step of solving, we will divide the whole equation by 100. After doing this, we will get the following equations:
\[\dfrac{100}{100}\times \dfrac{\sqrt{25}}{\sqrt{25}+x}=\dfrac{50}{100}\]
\[\Rightarrow \dfrac{\sqrt{25}}{\sqrt{25}+x}=\dfrac{1}{2}\]
Now, we will rationalize the above term on the left-hand side. By rationalization, we mean we will try to eliminate the irrational term in the denominator of the left-hand side of the equation. For doing this, we will multiply the numerator and denominator on the left-hand side with \[\sqrt{25}-x.\] Thus, after doing this, we will get,
\[\dfrac{\sqrt{25}}{\sqrt{25}+x}\times \dfrac{\sqrt{25}-x}{\sqrt{25}-x}=\dfrac{1}{2}\]
\[\Rightarrow \dfrac{\sqrt{25}\left( \sqrt{25}-x \right)}{\left( \sqrt{25}+x \right)\left( \sqrt{25}-x \right)}=\dfrac{1}{2}\]
In the above equation, we will use the following identity,
\[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]
Thus, we will get the following equation,
\[\Rightarrow \dfrac{\sqrt{25}\left( \sqrt{25}-x \right)}{{{\left( \sqrt{25} \right)}^{2}}-{{\left( x \right)}^{2}}}=\dfrac{1}{2}\]
\[\Rightarrow \dfrac{25-\sqrt{25}x}{25-{{x}^{2}}}=\dfrac{1}{2}\]
Now, we will cross multiply the terms. Thus, we will get,
\[\Rightarrow 2\left( 25-\sqrt{25}x \right)=25-{{x}^{2}}\]
\[\Rightarrow 2\left( 25-5x \right)=25-{{x}^{2}}\]
\[\Rightarrow 50-10x=25-{{x}^{2}}\]
\[\Rightarrow {{x}^{2}}-10x+50-25=0\]
\[\Rightarrow {{x}^{2}}-10x+25=0\]
\[\Rightarrow {{x}^{2}}-5x-5x+25=0\]
\[\Rightarrow x\left( x-5 \right)-5\left( x-5 \right)=0\]
\[\Rightarrow \left( x-5 \right)\left( x-5 \right)=0\]
\[\Rightarrow {{\left( x-5 \right)}^{2}}=0\]
\[\Rightarrow x=5\]
Now, we will use an exponential identity here,
\[a={{a}^{x\times \dfrac{1}{x}}}\]
In our case, a = 5 and we will put the value of x = 2. Thus we get,
\[\Rightarrow x={{5}^{2\times \dfrac{1}{2}}}\]
Now, we will use another exponential identity,
\[{{a}^{x\times y}}={{\left( {{a}^{x}} \right)}^{y}}\]
Thus we will get,
\[\Rightarrow x={{\left( {{5}^{2}} \right)}^{\dfrac{1}{2}}}\]
\[\Rightarrow x={{25}^{\dfrac{1}{2}}}\]
Now, we can write \[{{a}^{\dfrac{1}{2}}}\text{ as }\sqrt{a}.\] Thus, we will get,
\[\Rightarrow x=\sqrt{25}\]
Hence, the option (c) is correct.

Note: We can also solve this question alternatively by the following method,
\[\Rightarrow 100\times \dfrac{\sqrt{25}}{\sqrt{25}+x}=50\]
\[\Rightarrow \dfrac{\sqrt{25}}{\sqrt{25}+x}=\dfrac{1}{2}\]
\[\Rightarrow \dfrac{5}{5+x}=\dfrac{1}{2}\]
Now, we will take an inverse on both sides. After doing this, we will get,
\[\Rightarrow \dfrac{5+x}{5}=\dfrac{2}{1}\]
Now, we will subtract 1 from both sides. Thus, we will get,
\[\dfrac{5+x}{5}-1=2-1\]
\[\Rightarrow \dfrac{5+x-5}{5}=1\]
\[\Rightarrow \dfrac{x}{5}=1\]
\[\Rightarrow x=5\]
We can also write 5 as \[\sqrt{25}.\] So, we will get,
\[\Rightarrow x=\sqrt{25}\]