Solve for x: 1/((a + b + x)) = 1/a + 1/b + 1/x
[a β 0, b β 0, x β 0, x β β(a + b)]
CBSE Class 10 Sample Paper for 2019 Boards
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CBSE Class 10 Sample Paper for 2019 Boards
Last updated at Dec. 21, 2021 by Teachoo
Question 23 (OR 2nd question) Solve for x: 1/((π + π + π₯)) = 1/π + 1/π + 1/π₯ [a β 0, b β 0, x β 0, x β β(a + b)] 1/((π + π + π₯)) = 1/π + 1/π + 1/π₯ 1/((π + π + π₯)) β 1/π₯ = 1/π + 1/π (π₯ β (π + π + π₯))/(π + π + π₯)π₯ = 1/π + 1/π (π₯ β π β π β π₯)/(π + π + π₯)π₯ = (π + π)/ππ (β(π + π))/(π + π + π₯)π₯ = ((π + π))/ππ (β1)/(π + π + π₯)π₯ = 1/ππ βab = x(a + b + x) βab = x(a + b) + x2 0 = x2 + x(a + b) + ab x2 + x(a + b) + ab = 0 Comparing with Ax2 + BX + C = 0 A = 1, B = (a + b), C = ab x = (βπ΅ Β± β(π΅^(2 )β 4π΄πΆ) )/2π΄ x = (β(π + π) Β± β(γ(π + π)γ^(2 )β 4Γ1Γππ) )/(2 Γ 1) x = (β(π + π) Β± β(π^2 + π^2 + 2ππ β 4ππ) )/2 x = (β(π + π) Β± β(π^2 + π^2 β 2ππ) )/2 x = (β(π + π) Β± β((π β π)^2 ) )/2 x = (β(π + π) Β± (π β π) )/2 So, x = (β(π + π) + (π β π) )/2 x = (βπ β π + π β π)/2 x = (β2π)/2 x = βb x = (β(π + π) β (π β π) )/2 x = (βπ β π β π+ π)/2 x = (β2π)/2 x = βa