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If-a-gt-b-gt-0-and-4a-2-b-2-4ab-Find-a-b-a-b-

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Question Number 206477 by hardmath last updated on 15/Apr/24
If a>b>0 and 4a^2 + b^2 = 4ab Find: ((a − b)/(a + b)) = ?
$$\mathrm{If}\:\:\:\mathrm{a}>\mathrm{b}>\mathrm{0}\:\:\:\mathrm{and}\:\:\:\mathrm{4a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{4ab} \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{a}\:−\:\mathrm{b}}{\mathrm{a}\:+\:\mathrm{b}}\:=\:? \\ $$
Answered by MATHEMATICSAM last updated on 16/Apr/24
4a^2 + b^2 = 4ab ⇒ 4a^2 − 4ab + b^2 = 0 ⇒ (2a − b)^2 = 0 ⇒ 2a − b = 0 ⇒ 2a = b ⇒ (a/b) = (1/2) ⇒ ((a + b)/(a − b)) = ((1 + 2)/(1 − 2)) = − 3 [Comp. and Div.] ⇒ ((a − b)/(a + b)) = − (1/3) (Ans)
$$\mathrm{4}{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\mathrm{4}{ab} \\ $$$$\Rightarrow\:\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{4}{ab}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{2}{a}\:−\:{b}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}{a}\:−\:{b}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}{a}\:=\:{b} \\ $$$$\Rightarrow\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{{a}\:+\:{b}}{{a}\:−\:{b}}\:=\:\frac{\mathrm{1}\:+\:\mathrm{2}}{\mathrm{1}\:−\:\mathrm{2}}\:=\:−\:\mathrm{3}\:\left[\mathrm{Comp}.\:\mathrm{and}\:\mathrm{Div}.\right] \\ $$$$\Rightarrow\:\frac{{a}\:−\:{b}}{{a}\:+\:{b}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}}\:\left(\mathrm{Ans}\right) \\ $$
Answered by A5T last updated on 15/Apr/24
4a^2 −4ab+b^2 =0 ⇒a=((4b+_− (√(16b^2 −16b^2 )))/8)=(b/2) ⇒((a−b)/(a+b))=((−(b/2))/((3b)/2))=((−1)/3)
$$\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}{b}\underset{−} {+}\sqrt{\mathrm{16}{b}^{\mathrm{2}} −\mathrm{16}{b}^{\mathrm{2}} }}{\mathrm{8}}=\frac{{b}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{a}−{b}}{{a}+{b}}=\frac{−\frac{{b}}{\mathrm{2}}}{\frac{\mathrm{3}{b}}{\mathrm{2}}}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 15/Apr/24
4a^2 −4ab + b^2 = 0 (2a−b)^2 =0 b=2a ((a − b)/(a + b))=((a−2a)/(a+2a))=((−a)/(3a))=−(1/3)
$$\mathrm{4a}^{\mathrm{2}} −\mathrm{4ab}\:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\left(\mathrm{2a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{b}=\mathrm{2a} \\ $$$$\:\frac{\mathrm{a}\:−\:\mathrm{b}}{\mathrm{a}\:+\:\mathrm{b}}=\frac{\mathrm{a}−\mathrm{2a}}{\mathrm{a}+\mathrm{2a}}=\frac{−\mathrm{a}}{\mathrm{3a}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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