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Question Number 98073 by bobhans last updated on 11/Jun/20
solve (√(6−x)) = 6−x^2
solve6−x=6−x2
Commented by john santu last updated on 11/Jun/20
equation defined in 6−x≥0 ∧ 6−x^2 ≥0 ⇔ x ≤ 6 ∧ −(√6) ≤ x ≤ (√6) squaring 6−x = x^4 −12x^2 +36 x^4 −12x^2 +x+30 = 0 Horner method (x−2)(x+3)(x^2 −x−5) = 0 x = −3 (rejected) x = 2 (solution) x = ((1±(√(21)))/2) ⇒ { ((x=((1+(√(21)))/2) (rejected))),((x=((1−(√(21)))/2) (solution))) :}
equationdefinedin6−x⩾0∧6−x2⩾0⇔x⩽6∧−6⩽x⩽6squaring6−x=x4−12x2+36x4−12x2+x+30=0Hornermethod(x−2)(x+3)(x2−x−5)=0x=−3(rejected)x=2(solution)x=1±212⇒{x=1+212(rejected)x=1−212(solution)
Commented by bobhans last updated on 11/Jun/20
thank you both
thankyouboth
Answered by MJS last updated on 11/Jun/20
6−x=36−12x^2 +x^4 [squaring leads to false solutions] x^4 −12x^2 +x+30=0 (x−2)(x+3)(x^2 −x−5)=0 x_1 =2 x_2 =−3 x_(3, 4) =((1±(√(21)))/2) inserting in original equation gives x=2∨x=((1−(√(21)))/2)
6−x=36−12x2+x4[squaringleadstofalsesolutions]x4−12x2+x+30=0(x−2)(x+3)(x2−x−5)=0x1=2x2=−3x3,4=1±212insertinginoriginalequationgivesx=2∨x=1−212

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