Question

Find the value of k for which the quadratic equation
(k +1)x2 - 6(k + 1)x + 3(k + 9) = 0, k ≠-1 has equal root​

Answer 1

solution:

SOLUTION :  

Given : (p + 1)x² - 6(p + 1)x + 3(p + 9) = 0, p ≠ -1 has equal roots ………(1)

On comparing the given equation with ax² + bx + c = 0  

Here, a = p + 1 , b = - 6(p +1)  , c = 3(p +9)

D(discriminant) = b² – 4ac

D = [- 6(p +1)² - 4 × (p + 1) × 3(p + 9)

D = [36((p)² + 1²+ 2× p× 1)) - 12(p²  + 9p + p + 9)

[(a + b)² = a² + b² + 2ab]

D = 36(p² + 1 + 2p - 12(p²  + 10p + 9)

D = 36p² + 36 + 72p - 12 p² - 120 p - 108

D = 36p² - 12 p² + 72p - 120p + 36 - 108  

D = 24p² - 48p - 72

Given :  Equal roots  

Therefore , D = 0

24p² - 48p - 72  = 0

24(p² - 2p - 3) = 0  

p² - 2p - 3 = 0  

p² - 3p + p - 3 = 0  

[By middle term splitting]

p(p - 3) + 1 (p - 3) = 0

(p + 1) (p - 3) = 0

p + 1  = 0  or (p - 3) = 0

p = - 1  or p = 3

The value of p is - 1  & 3 .

It is given that p ≠ - 1 , so p = 3

Hence , the value of p is 3 only.

On putting p = 3 in eq 1 ,

(p + 1)x² - 6(p + 1)x + 3(p + 9) = 0

(3 + 1)x² - 6(3 + 1)x + 3(3 + 9) = 0

4x² - 6(4)x + 3(12) = 0

4x² - 24x + 36 = 0

4(x² - 6x + 9) = 0

x² - 6x + 9 = 0

x² - 3x  - 3x + 9 = 0

[By middle term splitting]

x(x - 3) -3(x - 3) = 0

(x - 3) = 0  or  (x - 3) = 0

x = 3  or  x = 3  

Roots are 3 & 3

Hence ,the roots of the equation (p + 1)x² - 6(p + 1)x + 3(p + 9) = 0 is 3 .

hope it will help you

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Value\:of\:k=-1\:and\:3}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k + 1)x^{2} -6( k + 1)x + 3(k+9) = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies ( k +1)x^{2} -(6k +6)x + (3k+27)= 0} \\ \\ \tt{\circ \: a = (k +1)} \\ \\ \tt{\circ \: b = -(6k + 6)}\\ \\\tt{\circ \:c = (3k+27)}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (-(6k +6))^{2} - 4 \times( k +1) \times (3k+27) = 0 } \\ \\ \tt{: \implies \: 36{k}^{2} + 36+72k - 12k^{2} -120k-108 = 0 } \\ \\ \tt{ : \implies \: 24{k}^{2} - 48k -72 = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} -2k - 3 = 0} \\ \\ \tt{: \implies k(k - 3) +1(k - 3) = 0} \\ \\ \tt{: \implies (k + 1)(k - 3) = 0} \\ \\ \green{\tt{: \implies k = -1 \: and \: 3}}$