Aspire Faculty ID #16173 · Topic: NIMCET 2011 · Just now
NIMCET 2011

Find least integer $k$ such that $(k-2)x^2 + k + 8x + 4 > 0$ for all $x\in\mathbb{R}$.

Solution

For quadratic $ax^2+bx+c>0$ for all $x$: $a>0$ → $k-2>0$ → $k>2$ Discriminant $<0$ $D=b^2-4ac=8^2-4(k-2)(k+4)$ Compute: $D=64-4(k^2+2k-8)=64-4k^2-8k+32$ $D=96-4k^2-8k<0$ Divide by $-4$: $k^2+2k-24>0$ $(k+?)(k+?)$ → roots $4$ and $-6$ So $k>4$ or $k<-6$ Combine with $k>2$ ⇒ $k>4$ Least integer = $5$

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