If - 4 is a root of the equation x^2 + 2x + 4p = 0, find the value of k for which the quadratic equation x^2 + px( 1 + 3k ) + 7 ( 3 + 2k ) = 0 has equal roots.
Answers
Given that - 4 is a root of the equation x^2 + 2x + 4p = 0
Therefore, x = - 4
Substituting the value of x in the given equation :-
⇒ x^2 + 2x + 4p = 0
⇒ ( - 4 )^2 + 2( - 4 ) + 4p = 0
⇒ 16 - 8 + 4p = 0
⇒ 8 + 4p = 0
⇒ 4p = - 8
⇒ p = - 8 / 4
⇒ p = - 2
Now,
Value of p is - 2. Substitute the value of p in the discriminant of equation x^2 + px( 1 + 3k ) + 7( 3 + 2k ) = 0
On the comparing the given equation x^2 + px( 1 + 3k ) + 7( 3 + 2k ) = 0 with ax^2 + bx + c = 0, we get that the following information :-
a = 1 , b = p( 1 + 3k ) , c = 7( 3 + 2k )
Discriminant of the quadratic equations = b^2 - 4ac
Given that the given ( 2nd ) equation has equal roots, so the result of discriminant should be equal to 0.
Now,
⇒ b^2 - 4ac = 0
⇒ [ p( 1 + 3k ) ]^2 - 4[ 1{ 7( 3 + 2k ) } ] =0
⇒ [ -2( 1 + 3k ) ]^2 - 28( 3 + 2k ) = 0
⇒ 4( 1 + 9k^2 + 6k ) - 4[ 7( 3 + 2k )] = 0
⇒ 1 + 9k^2 + 6k - 7( 3 + 2k ) = 0
⇒ 1 + 9k^2 + 6k - 21 - 14k = 0
⇒ 9k^2 - 8k - 20 = 0
⇒ 9k^2 + 10k - 18k - 20= 0
⇒ 9k^2 - 18k + 10k - 20 = 0
⇒ 9k( k - 2 ) + 10( k - 2 ) = 0
⇒ ( k - 2 )( 9k + 10 ) = 0
∴ k = 2 or k = - 10 / 9
Therefore the value of k is 2 or - 10 / 9.
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- 4 is a root of equation x'2 + 2x + 4p = 0
=> p(x)= p(-4) = 0
=> (-4)'2 + 2 * (-4) + 4p = 0
=> 16-8+4p= 0
=>4p= -8
=> p= -2...........................1
Given that quadratic equation
x '2 - px(1+3k) + 7(3+2k) = 0 has equal roots
=> discriminat b'2 - 4ac = 0
=> p²(1+3k)² - 4* 7(3+2k) = 0
since p = -2
=> 4 ( 1+3k)² - 28 ( 3+2k) = 0
=> 4( 1 + 6k + 9k²) - 84- 56k = 0
=> 4 + 24k + 36k² - 84 - 56k = 0
=> 36k² - 32k - 80 = 0
=> 9k² - 8k - 20 = 0
=> 9k² - 18k +10k -20 = 0
=> 9k(k - 2) + 10(k - 2) =0
=> (k-2) (9k + 10 ) = 0
=> k = 2 & -10/9