Menu Close

The-quadratic-equation-has-two-equal-roots-x-2-k-3-x-k-0-a-Find-the-value-of-k-b-For-this-value-of-k-solve-the-equation-for-x-c-If-x-is-the-length-of-a-rectangle-and-its-width-is-x-

Tinku Tara 0 Comments
Pin




Question Number 217058 by ArshadS last updated on 28/Feb/25
The quadratic equation has two equal roots: x^2 +(k−3)x+k=0 (a) Find the value of k. (b) For this value of k, solve the equation for x (c)If x is the length of a rectangle and its width is x−2, find the area of the rectangle.
$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:\:\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}: \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\left({k}−\mathrm{3}\right){x}+{k}=\mathrm{0} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{k}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{For}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:\:{k},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:{x} \\ $$$$\left(\mathrm{c}\right)\mathrm{If}\:\:{x}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{and}\:\mathrm{its}\:\mathrm{width}\:\mathrm{is}\:{x}−\mathrm{2},\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}. \\ $$
Answered by Rasheed.Sindhi last updated on 28/Feb/25
(a) For equal roots discriminant b^2 −4ac=0 (k−3)^2 −4(1)(k)=0 k^2 −10k+9=0⇒k=1,9 (b) •k=1: x^2 +(k−3)x+k=0 ⇒x^2 −2x+1=0⇒x=1 •k=9: x^2 +6k+9=0⇒x=−3 (c) •x=1 length of rectangle=x=1 width of rectangle=1−2=−1 Impossible •x=−3: aea is impssible. area is impossibe in both cases (i-e There is no rectangle)
$$\left(\mathrm{a}\right) \\ $$$${For}\:{equal}\:{roots}\:{discriminant} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0} \\ $$$$\left({k}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left({k}\right)=\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{10}{k}+\mathrm{9}=\mathrm{0}\Rightarrow{k}=\mathrm{1},\mathrm{9} \\ $$$$\left(\mathrm{b}\right) \\ $$$$\bullet{k}=\mathrm{1}: \\ $$$$\:{x}^{\mathrm{2}} +\left({k}−\mathrm{3}\right){x}+{k}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$$\bullet{k}=\mathrm{9}: \\ $$$$\:\:{x}^{\mathrm{2}} +\mathrm{6}{k}+\mathrm{9}=\mathrm{0}\Rightarrow{x}=−\mathrm{3} \\ $$$$\left(\mathrm{c}\right) \\ $$$$\bullet{x}=\mathrm{1} \\ $$$${length}\:{of}\:{rectangle}={x}=\mathrm{1} \\ $$$${width}\:{of}\:{rectangle}=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$$${Impossible} \\ $$$$\bullet{x}=−\mathrm{3}:\:{aea}\:{is}\:{impssible}. \\ $$$${area}\:{is}\:{impossibe}\:{in}\:{both}\:{cases} \\ $$$$\left({i}-{e}\:{There}\:{is}\:{no}\:{rectangle}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *