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Ask QuestionPosted by ☺️☺️☺️? ..?? 6 years, 3 months ago
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#Aditi~ Angel???? 6 years, 3 months ago
Posted by Harshita Sharma?? 6 years, 3 months ago
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Posted by Aayush Kamal 6 years, 3 months ago
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Posted by Aayush Kamal 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
We have, (k+4) x2 + (k+1)x + 1= 0
Here a = (k+4), b = (k+1), c =1
{tex}\implies{/tex}D = b2 -4ac = (k+1)2 - 4 (k+4) (1)
= k2 +1 + 2k - 4k -16= k2 -2k -15
For equal roots, D = 0
{tex}\implies{/tex}k2 - 2k -15 = 0
{tex}\implies{/tex}k2 - 5k + 3k -15 = 0
{tex}\implies{/tex}k (k - 5) +3 (k - 5) = 0
{tex}\implies{/tex}(k+3) (k-5) = 0
Either k+3 = 0 or k-5 = 0
{tex}\implies{/tex}k = -3, 5
Posted by Pratik Dhonde 6 years, 3 months ago
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Khushi Keshri 6 years, 3 months ago
Posted by Yeshiv Bhasin 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
If g(x) = x2 + 2x + k is a factor of f(x) = 2x4 + x3 - 14x2 + 5x + 6, then remainder is zero when f(x) is divided by g(x).
Let quotient = Q and remainder = R
Let us now divide f(x) by g(x).
R = x(7k + 21) + (2k2 + 8k + 6) -------(1) and Q = 2x2 - 3x - 2(k + 4).------------(2)
Now, R = 0.
{tex}\Rightarrow{/tex} x (7k + 21) + 2 (k2 + 4k + 3) = 0
{tex}\Rightarrow{/tex} 7x (k + 3) + 2 (k+1)(k+3) = 0
{tex}\Rightarrow{/tex} (k+3) [7x + 2(k+1)] = 0
{tex}\Rightarrow{/tex} k + 3 = 0
{tex}\Rightarrow{/tex} k = -3
Thus, polynomial f(x) can be written as,
2x4 + x3 - 14x2 + 5x + 6 = (x2 + 2x + k) [2x2 - 3x - 2(k + 4)] = (x2 + 2x - 3) (2x2 - 3x - 2)
Zeros of x2 + 2x - 3 are,
x2 + 2x - 3 = 0
{tex}\Rightarrow{/tex} (x + 3) (x - 1) = 0
{tex}\Rightarrow{/tex} x = -3 or x = 1
Zeros of (2x2 - 3x - 2) are,
2x2 - 3x - 2 = 0
{tex}\Rightarrow{/tex} 2x2 - 4x + x - 2 = 0
{tex}\Rightarrow{/tex} 2x(x - 2) + 1(x - 2) = 0
{tex}\Rightarrow{/tex} (x - 2)(2x + 1) = 0
x = 2 or x = -{tex}\frac12{/tex}
Thus, the zeros of f(x) are: -3 ,1, 2 and -{tex}\frac12{/tex}
Posted by Vikram Gujjar 6 years, 3 months ago
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Shravan Kumar 6 years, 3 months ago
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Sourav Sharma 6 years, 3 months ago
Gourav ## Back ## Happy ? ? 6 years, 3 months ago
Posted by Rahul Ray 6 years, 3 months ago
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Posted by Rahul Ray 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Let the actual age of Zeba be 'x' years. If her age is 5 years less than the actual, her age would be (x-5) years. Clearly; x ≥5......(1)
According to the question, we have
(x - 5)2 = 5x + 11
{tex}\Rightarrow{/tex} x2 - 10x + 25 = 5x + 11
{tex}\Rightarrow{/tex} x2 - 15x + 14 = 0
{tex}\Rightarrow{/tex} x2 - 14x - x + 14 = 0
{tex}\Rightarrow{/tex} (x - 14) - 1(x - 14) = 0
{tex}\Rightarrow{/tex} (x - 14)(x - 1) = 0
{tex}\Rightarrow{/tex} x - 14 = 0 or x - 1 = 0
{tex}\Rightarrow{/tex} x = 14 or x = 1
Since x > 5, we have x = 14
Thus, the present age of Zeba is 14 years.
Posted by Sharique Ali 6 years, 3 months ago
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Posted by Pramisi Singh 6 years, 3 months ago
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Ramesh Kumar 6 years, 3 months ago
Posted by Pramila Kaushik 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Let the ratio be K: 1
Coordinate of P are {tex}\left( {\frac{{3K + 2}}{{K + 1}},\frac{{7K - 2}}{{K + 1}}} \right){/tex}
P lies on the line 2x + y - 4 = 0
{tex} \Rightarrow 2\left( {\frac{{3K + 2}}{{K + 1}}} \right) + \frac{{7K - 2}}{{K + 1}} - \frac{4}{1} = 0{/tex}
{tex} \Rightarrow {/tex} 6K + 4 + 7K - 2 - 4K - 4 = 0
{tex} \Rightarrow {/tex} 9K - 2 = 0
{tex} \Rightarrow K = \frac{2}{9}{/tex} or 2 : 9
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Neeraj Kumar 6 years, 3 months ago
0Thank You