Question

this is from ch 4 of class 10 th math ​

Answer 1

Given two quadratic equations,

3x² + 7x + p = 0 ---(1)

and

x² + k(4x+k-1)+p = 0 ---(2)

-2 is a common root of the equations.

/* substitute x = -2 in the equations we get*/

i ) 3(-2)² + 7(-2) + p = 0

=> 3 × 4 - 14 + p = 0

=> 12 - 14 + p = 0

=> -2 + p = 0

=> p = 2 ---(3)

ii ) Now , substitute x = -2 and p = 2 in equation (2), we get

(-2)² + k[ 4(-2) + k - 1 ] + 2 = 0

=> 4 + k( -8 + k - 1 ) + 2 = 0

=> 4 + k( k - 9 ) + 2 = 0

=> 4 + k² - 9k + 2 = 0

=> k² - 9k + 6 = 0

Compare this with ak² + bk + c = 0 , we get

a = 1 , b = -9, c = 6

Discreminant (D) = b² - 4ac

= (-9)² - 4×1×6

= 81 - 24

= 57

By Quadratic Formula:

$k = \frac{ -b±\sqrt{D}}{2a}$

$= \frac{ -(-9)±\sqrt{57}}{2\times 1 }\\= \frac{ 9±\sqrt{57}}{2 }$

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