CBSE 2025 · Region 3 · Set 1 · Q24 · 2 marks
Find the value(s) of 'k' so that the quadratic equation $\displaystyle 4 \mathrm{x}^{2}+\mathrm{kx}+1=0$ has real and equal roots.If ' $\displaystyle \alpha$ ' and ' $\displaystyle \beta$ ' are the zeroes of the polynomial $\displaystyle \mathrm{p}(\mathrm{y})=\mathrm{y}^{2}-5 \mathrm{y}+3$, then find the value of $\displaystyle \alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}$.
Find the value(s) of 'k' so that the quadratic equation $\displaystyle 4 \mathrm{x}^{2}+\mathrm{kx}+1=0$ has real and equal roots.
If ' $\displaystyle \alpha$ ' and ' $\displaystyle \beta$ ' are the zeroes of the polynomial $\displaystyle \mathrm{p}(\mathrm{y})=\mathrm{y}^{2}-5 \mathrm{y}+3$, then find the value of $\displaystyle \alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}$.
Marking-scheme solution
(a)
For real and equal roots, \(\displaystyle \mathrm{D}=0\)
\[\begin{aligned}
& \mathrm{k}^{2}-16=0 \\
& \mathrm{k}= \pm 4
\end{aligned}
\]
(b)
\[\begin{aligned}
& \alpha+\beta=5 \\
& \alpha \beta=3 \\
& \alpha^{4} \beta^{3}+\alpha^{3} \beta^{4}=(\alpha \beta)^{3}(\alpha+\beta) \\
& =27 \times 5=135
\end{aligned}
\]
Quadratic EquationsNature of Roots and the DiscriminantApplyvery_short_answermedium
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