Question

Find out the value of k _____

Where given that a polynomial has equal roots


$(k+1)x²+(7k+7)x+4K+9$
​

Answer 1

Given :-

• A polynomial __

(k+1)x² + (7k+7)x + 4k+9

Where, roots are equal.

To Find :-

• Value of k.

Formula to be used :-

• $\bold{\big[\:(a+b)^2\:=\:a^2\:+\:2ab\:+\:b^2\big]}$

Solution :-

Given that,

a polynomial __(k+1)x² + (7k+7)x + 4k+9

Where,

a = (k+1)

b = (7k + 7)

c = 4k + 9

We know that, a polynomial has equal roots only if the discriminant,D equals to zero.

We know,

$\bold{D\:=\:b^2\:-\:4ac}$

Now, put the acquired values

$\sf{\longrightarrow{0\:=\:(7k+7)^2-4\big[(k+1)(4k+9)}\big]}$

$\sf{\longrightarrow{0\:=\:(7k) ^2\:+\:(2\:\times\:7k\:\times\:7)\:+\:(7)^2\:-\:4\:\times\:\big[k(4k+9)+1(4k+9)\big]}}$

$\sf{\longrightarrow{0\:=\:49k^2\:+\:(14k\:\times\:7\:)\times\:49\:-\:4\:\times\:\big[4k^2+9k\:+\:4k+9}\big]}$

$\sf{\longrightarrow{0\:=\:49k^2\:+\:98k\:+\:49\:-\:4\:\times\:\big[4k^2+13k+9}\big]}$

$\sf{\longrightarrow{0\:=\:49k^2+98k+49\:-\:16k^2-52k-36}}$

$\sf{\longrightarrow{0\:=\:49k^2-16k^2+98k-52k+49-36}}$

$\sf{\longrightarrow{33k^2+46k+13=0}}$

_____________________________________________________

Now, we need to use factorization method for solving this equation.

$\sf{\longrightarrow{33k^2+33k+13k+13=0}}$

$\sf{\longrightarrow{33k(k+1)+13(k+1)=0}}$

$\sf{\longrightarrow{(k+1)(33k+13)=0}}$

$\sf{\longrightarrow{k+1=0\:\:or\:\:33k+13=0}}$

$\sf{\longrightarrow{k=-1\:\:or\:\:33k=-13}}$

$\sf{\longrightarrow{k=-1\:\:or\:\:k=\dfrac{-13}{33}}}$

Hence,

$</p><p></p><p>\tt\green{Value\:of\:k\:=\:-1\:or\:\dfrac{-13}{33}}$