Quadratic Equations

### Show the is a solution of

The given equation is $x^{2}+6 x+9=0$ Putting $x=-3$ in the given equation, we get $L H S=(-3)^{2}+6 \times(-3)+9=9-18+9=0=R H S$ $\therefore x=-3$ is a solution of the given equation.

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### A train covers a distance of at a uniform speed. If the speed had been less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.

Let the usual speed of the train be $x \mathrm{~km} / \mathrm{h}$. $\therefore$ Reduced speed of the train $=(x-8) \mathrm{km} / \mathrm{h}$ Total distance to be covered $=480 \mathrm{~km}$ Time...

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### Find the roots of the given equation:

$\begin{array}{l} x^{2}-6 x+3=0 \\ \Rightarrow x^{2}-6 x=-3 \\ \Rightarrow x^{2}-2 \times x \times 3+3^{2}=-3+3^{2} \\ \Rightarrow(x-3)^{2}=-3+9=6 \\ \Rightarrow x-3=\pm \sqrt{6} \end{array}$...

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### 2. A train, traveling at a uniform speed for km, would have taken minutes less to travel the same distance if its speed were km/hr more. Find the original speed of the train.

Solution: Let the original speed of train be x km/hr When increased by $5$, speed of the train $= (x + 5) km/hr$ Using, speed = distance/...

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### 1. The speed of a boat in still water is km/hr. It can go km upstream and km downstream in hours. Find the speed of the stream.

Solution: Let the speed of stream be x km/hr Given, speed of boat in still water is $8km/hr$. So, speed of downstream $= (8 + x) km/hr$ And, speed of upstream $= (8 – x) km/hr$ Using, speed =...

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