Question

if x is equals to minus 2 is a roots of the equation 3 X square + 7 x + p equals to zero find K if root of equation are equal X square + k (4 x + K - 1 )+ p is equals to zero​

Answer 1

➫QuesTion :-

If x = -2 is a root of the equation 3x² + 7x + p = 0 ,find the value of k is roots of the equation

x² + k(4x + k - 1) + p = 0 .

➫Answer :-

$\: \: \to \: \boxed{ \sf \: k \: = - 1 \: or \: \frac{2}{3} } \:$

➫To Find :-

→ Value of k .

➫Solution :-

According to the question,

→ -2 is a roof of equation 3x² + 7x + p = 0

let , f(x) = 3x² + 7x + p = 0

if -2 is a root of this equation then it will satisfy to this Quadratic ,

→ f(-2) = 3 (-2)² + 7(-2) + p = 0

→ 3×4 - 14 + p = 0

→ 12 - 14 + p = 0

→ -2+ p = 0

→ p = 2

We have another Quadratic whose roots are equal .

→ g(x) = x² + 4kx + k² - k + p =0

We have p = 2

→ g(x) = x² + 4kx + k² - k + 2 = 0

• ° • we know that if roots of any Quadratic equation are equal then its Discriminant is 0

here ,

→ a = 1 , b = 4k and c = k² - k + 2

• ° • Discriminant (D) = b² - 4ac = 0

→ (4k)² - 4 × 1 × (k² - k + 2) = 0

→ 16k² - 4k² + 4k - 8 = 0

→ 12k² + 4k - 8 = 0

→ 4( 3k² + k - 2) = 0

Splitting the middle term ,

→ 3k² + 3k - 2k - 2 = 0

→ 3k ( k + 1) -2( k + 1) = 0

→( k + 1)(3k - 2) = 0

$\star \: \tt case \: (1) \: if \: \\ \to \tt \: k \: + 1 = 0 \\ \\ \to \boxed{ \tt \: k \: = - 1} \\ \\ \star \tt \: case \: (2) \: if \\ \\ \to \tt \: 3k \: - 2 = 0 \\ \\ \to \: \tt 3k = 2 \\ \\ \to \: \boxed{ \tt k = \frac{2}{3} }$

Hence ,

$\boxed{ \sf \: k \: = - 1 \: or \: \frac{2}{3} }$