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Question Number 94874 by i jagooll last updated on 21/May/20

Commented by hknkrc46 last updated on 21/May/20

(√(2−7x))+2x=0 ★ ((f(x)))^(1/(2n)) ≥0 ∧ f(x)≥0 ; n∈Z^+ ★ n=1 ⇒ ((f(x)))^(1/2) =(√(f(x))) ⇒ (√(2−7x))=−2x (1) 2−7x≥0 ⇒ 2≥7x ⇒ x≤(2/7) (2) (√(2−7x))=−2x≥0 ⇒ x≤0 ✠ x≤(2/7) ∧ x≤0 ⇒ x≤0 (3) ((√(2−7x)))^2 =(−2x)^2 ⇒ 2−7x=4x^2 ⇒ 4x^2 +7x−2=0 ★ ax^2 +bx+c=0 ⇒ x_(1,2) =((−b∓(√(b^2 −4ac)))/(2a)) x_1 =((−b−(√(b^2 −4ac)))/(2a))=((−7−(√(7^2 −4∙4∙(−2))))/(2∙4))=−2 x_2 =((−b+(√(b^2 −4ac)))/(2a))=((−7+(√(7^2 −4∙4∙(−2))))/(2∙4))=(1/4) {x_1 =−2 < 0 ∧ x_2 =(1/4) ≮ 0} (√(2−7x))+2x=0 ⇒ x=−2

2−7x+2x=0★f(x)2n⩾0∧f(x)⩾0;n∈Z+★n=1⇒f(x)2=f(x)⇒2−7x=−2x(1)2−7x⩾0⇒2⩾7x⇒x⩽27(2)2−7x=−2x⩾0⇒x⩽0✠x⩽27∧x⩽0⇒x⩽0(3)(2−7x)2=(−2x)2⇒2−7x=4x2⇒4x2+7x−2=0★ax2+bx+c=0⇒x1,2=−b∓b2−4ac2ax1=−b−b2−4ac2a=−7−72−4⋅4⋅(−2)2⋅4=−2x2=−b+b2−4ac2a=−7+72−4⋅4⋅(−2)2⋅4=14{x1=−2<0∧x2=14≮0}2−7x+2x=0⇒x=−2

Answered by john santu last updated on 21/May/20

(√(2−7x)) = −2x , for x < 0 squaring in both sides 2−7x = 4x^2 ⇒4x^2 +7x−2=0 (4x−1)(x+2) = 0 { ((x=−2 ( solution))),((x=(1/4) (not solution))) :}

2−7x=−2x,forx<0squaringinbothsides2−7x=4x2⇒4x2+7x−2=0(4x−1)(x+2)=0{x=−2(solution)x=14(notsolution)

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