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Question Number 217058 by ArshadS last updated on 28/Feb/25

$$\mathrm{The}\mathrm{quadratic}\mathrm{equation}\mathrm{has}\mathrm{two}\mathrm{equal}\mathrm{roots}:$$$$x^2+(k-3)x+k=0$$$$\left(\mathrm{a}\right)\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}k.$$$$\left(\mathrm{b}\right)\mathrm{For}\mathrm{this}\mathrm{value}\mathrm{of}k,\mathrm{solve}\mathrm{the}\mathrm{equation}\mathrm{for}x$$$$\left(\mathrm{c}\right)\mathrm{If}x\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{and}\mathrm{its}\mathrm{width}\mathrm{is}x-2,\mathrm{find}\mathrm{the}\mathrm{area}$$$$\mathrm{of}\mathrm{the}\mathrm{rectangle}.$$

Answered by Rasheed.Sindhi last updated on 28/Feb/25

$$\left(\mathrm{a}\right)$$$$Forequalrootsdiscriminant$$$$b^2-4ac=0$$$${(k-3)}^2-4\left(1\right)\left(k\right)=0$$$$k^2-10k+9=0\Rightarrow k=1,9$$$$\left(\mathrm{b}\right)$$$$\bullet k=1:$$$$x^2+(k-3)x+k=0$$$$\Rightarrow x^2-2x+1=0\Rightarrow x=1$$$$\bullet k=9:$$$$x^2+6k+9=0\Rightarrow x=-3$$$$\left(\mathrm{c}\right)$$$$\bullet x=1$$$$lengthofrectangle=x=1$$$$widthofrectangle=1-2=-1$$$$Impossible$$$$\bullet x=-3:aeaisimpssible.$$$$areaisimpossibeinbothcases$$$$(i-eThereisnorectangle)$$

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