Solve \((8-t)^2 < \left(t^2-3 t-10\right)\)
Solve \((8-t)^2 < \left(t^2-3 t-10\right)\)
- \(\left(\frac{74}{13}, 8\right]\)
- \(\left(\frac{74}{13}, \infty\right)\)
- \((8, \infty)\)
- \([8, \infty)\)
Solution
It is given that,
\(\begin{array}{rlrl}
& & (8-t)^2 < t^2-3 t-10 \\
\Rightarrow & 64-16 t+t^2 < t^2-3 t-10 \\
\Rightarrow & 13 t > 74 \\
\Rightarrow & t > \frac{74}{13} \\
\Rightarrow & t \in\left(\frac{74}{13}, \infty\right)
\end{array}\)
Hence, option (b) is correct.
Asked in: AP EAMCET 2020 (21 Sep Shift 2)
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