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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 5 · Set 1 · Q32
(a) Find the value of 'k' for which the quadratic equation ( k +1) x^2-6( k +1) x +3( k +9)=0, k ≠-1 has real and equal roots.
(a)
Find the value of 'k' for which the quadratic equation $\displaystyle (\mathrm{k}+1) \mathrm{x}^{2}-6(\mathrm{k}+1) \mathrm{x}+3(\mathrm{k}+9)=0, \mathrm{k} \neq-1$ has real and equal roots.
OR
(b)
The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be $\displaystyle 4$ years more than three times the age of his son. Find their present ages.
Marking-scheme solution
(a)
For real and equal roots, \(\displaystyle \mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac}=0\)
\(\displaystyle 36(\mathrm{k}+1)^{2}-4(\mathrm{k}+1) \times 3(\mathrm{k}+9)=0\)
\(\displaystyle \Rightarrow \mathrm{k}^{2}-2 \mathrm{k}-3=0\)
\(\displaystyle \Rightarrow(\mathrm{k}-3)(\mathrm{k}+1)=0\)
\(\displaystyle \mathrm{k} \neq-1\) So, \(\displaystyle \mathrm{k}=3\)
OR
(b)
Let present age of son = x years
and present age of man \(\displaystyle =2 \mathrm{x}^{2}\) years
A.T.Q.
\(\displaystyle 3(x+8)+4=2 x^{2}+8\)
\[\begin{aligned}
& \Rightarrow 2 x^{2}-3 x-20=0 \\
& \Rightarrow(2 x+5)(x-4)=0 \\
& x \neq \frac{5}{2} S o, x=4
\end{aligned}
\]
Present age of son = $\displaystyle 4$ years
Present age of man = $\displaystyle 32$ years
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