Question

\( \begin{array}{l}\text { i. } x^{2}+14 x+49>0 \\ 2 \cdot x^{2}+20 x+16 \leq 0 \\ 3 \cdot x^{2}-8 x+16 \leq 0 \\ 1 \cdot x^{2}-10 x+29>0 \\ 5 \cdot x^{2}-x-20 \geq 0\end{array} \)

          \( \begin{array}{l}\text { i. } x^{2}+14 x+49>0 \\ 2 \cdot x^{2}+20 x+16 \leq 0 \\ 3 \cdot x^{2}-8 x+16 \leq 0 \\ 1 \cdot x^{2}-10 x+29>0 \\ 5 \cdot x^{2}-x-20 \geq 0\end{array} \)

i. x^2+14 x+49>0
2 · x^2+20 x+16 ≤ 0
3 · x^2-8 x+16 ≤ 0
1 · x^2-10 x+29>0
5 · x^2-x-20 ≥ 0

Added by Victoria T.

Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Sanjiva Dayal IIT Kanpur 1st Edition

Chapter 7

Instant Answer

Solved by Expert Nick Johnson

12/02/2022

Step 1

The quadratic expression \(x^2 + 14x + 49\) can be factored as \((x + 7)^2\). Since \((x + 7)^2\) is always non-negative and equals zero when \(x = -7\), the inequality \((x + 7)^2 > 0\) holds for all \(x \neq -7\). So, the solution is \(x \in (-\infty, -7) Show more…

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\( \begin{array}{l}\text { i. } x^{2}+14 x+49>0 \\ 2 \cdot x^{2}+20 x+16 \leq 0 \\ 3 \cdot x^{2}-8 x+16 \leq 0 \\ 1 \cdot x^{2}-10 x+29>0 \\ 5 \cdot x^{2}-x-20 \geq 0\end{array} \)