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If \( m\) and \( n\)\(\left( m > n\right) \) are the roots of the equation
\( 7\)\( \left( x + 2a\right)^{2}\) + \( 3a^{2}\) = \( 5a\)\( \left( 7x + 23a\right)\)
Where \( a>0\) then what is \( 3m\) - \( n\) equal to ?
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A) 12a |
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B) 14a |
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C) 15a |
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D) 18a |
Correct Answer:
| C) 15a |
Description for Correct answer:
Here a > 0
Let us consider a = 1,
\( \Large 7(x + 2a)^{2} + 3a^{2} = 5a(7x + 23a)\)
\( \Large 7(x + 2)^{2} + 3 = 5(7x + 23) \)
\( \Large 7(x^{2} + 4x +4) + 3 = 35x + 115 \)
\( \Large 7x^{2} + 28x - 35x + 28 + 3 - 115 = 0 \)
\( \Large 7x^{2} - 7x - 84 =0 \)
Now, m + n = 1 ...(i)
and mn = - 12
Now \( \Large (m-n)^{2} = (m+n)^{2} - 4mn \)
= 1-4(-12)=49
m - n = + 7 ... (ii)
Now solving equation (i) and (ii) we get
For m = 4
n = - 3
and m = 3, n = + 4
for m = 4 and n = - 3
Now 3m -n = 3(4) + 3 = 15
Option (C)is satisfied Option (C)is correct.
Here a > 0
Let us consider a = 1,
\( \Large 7(x + 2a)^{2} + 3a^{2} = 5a(7x + 23a)\)
\( \Large 7(x + 2)^{2} + 3 = 5(7x + 23) \)
\( \Large 7(x^{2} + 4x +4) + 3 = 35x + 115 \)
\( \Large 7x^{2} + 28x - 35x + 28 + 3 - 115 = 0 \)
\( \Large 7x^{2} - 7x - 84 =0 \)
Now, m + n = 1 ...(i)
and mn = - 12
Now \( \Large (m-n)^{2} = (m+n)^{2} - 4mn \)
= 1-4(-12)=49
m - n = + 7 ... (ii)
Now solving equation (i) and (ii) we get
For m = 4
n = - 3
and m = 3, n = + 4
for m = 4 and n = - 3
Now 3m -n = 3(4) + 3 = 15
Option (C)is satisfied Option (C)is correct.
Part of solved CDS Maths(1) questions and answers : Exams >> CDSE >> CDS Maths(1)
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