Mathematics · 2021
JEE Main · 24 February 2021, Shift 2 · Q90
The number of the real roots of the equation (x+1)^2+|x-5|=27/4 is \ \ \ \.
The number of the real roots of the equation $\displaystyle (x+1)^2+|x-5|=\frac{27}{4}$ is \_\_\_\_.
Official answer
2
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.