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Mathematics · 2022 · 2 marks
CBSE 2022 · Region 4 · Set 3 · Q1
Find the value of 'k' for which the quadratic equation $\displaystyle 2 \mathrm{k} x^{2}-40 x+25=0$ has real and equal roots.Solve for $\displaystyle x$ : $\displaystyle \frac{5}{2} x^{2}+\frac{2}{5}=1-2 x$.
Find the value of 'k' for which the quadratic equation $\displaystyle 2 \mathrm{k} x^{2}-40 x+25=0$ has real and equal roots.
Solve for $\displaystyle x$ : $\displaystyle \frac{5}{2} x^{2}+\frac{2}{5}=1-2 x$.
Marking-scheme solution
\[\begin{aligned}
& 2 k x^{2}-40 x+25=0 \\
& a=2 \mathrm{k}, b=-40, c=25 \\
& \text { Q.E. has equal roots } \Rightarrow D=0 \\
& b^{2}-4 a c=0 \Rightarrow 1600-200 k=0 \\
& k=8
\end{aligned}
\]
Or
\[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-20 \pm \sqrt{400+600}}{2 \times 25}
\]
\[x=\frac{-20 \pm \sqrt{1000}}{50}=\frac{-20 \pm 10 \sqrt{10}}{50}
\]
\[x=\frac{-2 \pm \sqrt{10}}{5} ; x=\frac{-2+\sqrt{10}}{5}, \frac{-2-\sqrt{10}}{5}
\]
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