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x-2-5x-6-0-amp-x-2-kx-1-0-have-a-common-root-then-k-




Question Number 205021 by BaliramKumar last updated on 06/Mar/24
x^2  + 5x +6 = 0 & x^2  + kx + 1 = 0 have a   common root then  k = ?
$${x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\mathrm{6}\:=\:\mathrm{0}\:\&\:{x}^{\mathrm{2}} \:+\:{kx}\:+\:\mathrm{1}\:=\:\mathrm{0}\:{have}\:{a}\: \\ $$$${common}\:{root}\:\mathrm{then}\:\:{k}\:=\:? \\ $$
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) ✓
$${x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\mathrm{6}\:=\:\mathrm{0}\:\&\:{x}^{\mathrm{2}} \:+\:{kx}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\wedge\begin{cases}{{x}^{\mathrm{2}} +{kx}+\mathrm{1}=\left({x}+\mathrm{2}\right)\left({x}+{p}\right)=\mathrm{0}}\\{{x}^{\mathrm{2}} +{kx}+\mathrm{1}=\left({x}+\mathrm{3}\right)\left({x}+{q}\right)=\mathrm{0}}\end{cases}\: \\ $$$$\bullet{x}^{\mathrm{2}} +\left({p}+\mathrm{2}\right){x}+\mathrm{2}{p}=\mathrm{0} \\ $$$$\:\:{p}+\mathrm{2}={k}\:\wedge\:\mathrm{2}{p}=\mathrm{1}\Rightarrow{k}=\frac{\mathrm{5}}{\mathrm{2}}\:\checkmark \\ $$$$\bullet{x}^{\mathrm{2}} +\left({q}+\mathrm{3}\right){x}+\mathrm{3}{q}=\mathrm{0} \\ $$$$\:\:\:{q}+\mathrm{3}={k}\:\wedge\:\mathrm{3}{q}=\mathrm{1}\Rightarrow{k}=\frac{\mathrm{10}}{\mathrm{3}}\:\checkmark \\ $$
Commented by BaliramKumar last updated on 06/Mar/24
thanks
$$\mathrm{thanks} \\ $$
Answered by mr W last updated on 06/Mar/24
roots of x^2 +5x+6=0 are x=−2, −3.  so we have  (−2)^2 +k(−2)+1=0 ⇒k=(5/2) or  (−3)^2 +k(−3)+1=0 ⇒k=((10)/3)
$${roots}\:{of}\:{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}=\mathrm{0}\:{are}\:{x}=−\mathrm{2},\:−\mathrm{3}. \\ $$$${so}\:{we}\:{have} \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{2}} +{k}\left(−\mathrm{2}\right)+\mathrm{1}=\mathrm{0}\:\Rightarrow{k}=\frac{\mathrm{5}}{\mathrm{2}}\:{or} \\ $$$$\left(−\mathrm{3}\right)^{\mathrm{2}} +{k}\left(−\mathrm{3}\right)+\mathrm{1}=\mathrm{0}\:\Rightarrow{k}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$
Commented by BaliramKumar last updated on 06/Mar/24
thanks
$$\mathrm{thanks} \\ $$

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