x^2 + 5x +6 = 0 & x^2 + kx + 1 = 0 have a common root then k = ?


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Question Number 205021 by BaliramKumar last updated on 06/Mar/24

$x^{2} + 5 x + 6 = 0 & x^{2} + k x + 1 = 0 h a v e a$ $c o m m o n r o o t then k = ?$
	Answered by Rasheed.Sindhi last updated on 06/Mar/24
	$x^2 + 5x +6 = 0 & x^2 + kx + 1 = 0 (x+2)(x+3)=0 ∧ { ((x^2 +kx+1=(x+2)(x+p)=0)),((x^2 +kx+1=(x+3)(x+q)=0)) :} •x^2 +(p+2)x+2p=0 p+2=k ∧ 2p=1⇒k=(5/2) ✓ •x^2 +(q+3)x+3q=0 q+3=k ∧ 3q=1⇒k=((10)/3) ✓$
	$x^{2} + 5 x + 6 = 0 & x^{2} + k x + 1 = 0$ $(x + 2) (x + 3) = 0$ $\land {\begin{cases} x^{2} + k x + 1 = (x + 2) (x + p) = 0 \\ x^{2} + k x + 1 = (x + 3) (x + q) = 0 \end{cases}$ $∙ x^{2} + (p + 2) x + 2 p = 0$ $p + 2 = k \land 2 p = 1 \Rightarrow k = \frac{5}{2} ✓$ $∙ x^{2} + (q + 3) x + 3 q = 0$ $q + 3 = k \land 3 q = 1 \Rightarrow k = \frac{10}{3} ✓$
	Commented by BaliramKumar last updated on 06/Mar/24

	$thanks$
	Answered by mr W last updated on 06/Mar/24

	$r o o t s o f x^{2} + 5 x + 6 = 0 a r e x = - 2, - 3 .$ $s o w e h a v e$ ${(- 2)}^{2} + k (- 2) + 1 = 0 \Rightarrow k = \frac{5}{2} o r$ ${(- 3)}^{2} + k (- 3) + 1 = 0 \Rightarrow k = \frac{10}{3}$
	Commented by BaliramKumar last updated on 06/Mar/24

	$thanks$
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