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Question Number 141669 by bramlexs22 last updated on 22/May/21

Answered by iloveisrael last updated on 22/May/21

(i) f(a)=g(a) ,a>0 ⇒ a+∣2a∣ = −a^2 −(2/3)a+((14)/3) ⇒3a =−a^2 −(2/3)a+((14)/3) ⇒3a^2 +9a+2a−14=0 ⇒3a^2 +11a−14=0 (ii) f(c)=g(c); c<0 ⇒−c=−c^2 −(2/3)c+((14)/3) ⇒3c^2 −3c+2c−14=0 ⇒3c^2 −c−14=0

(i)f(a)=g(a),a>0⇒a+∣2a∣=−a2−23a+143⇒3a=−a2−23a+143⇒3a2+9a+2a−14=0⇒3a2+11a−14=0(ii)f(c)=g(c);c<0⇒−c=−c2−23c+143⇒3c2−3c+2c−14=0⇒3c2−c−14=0

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