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Question Number 31915 by rahul 19 last updated on 16/Mar/18

If : (x^2 +x+2)^2 −(a−3)(x^2 +x+1)(x^2 +x+2) + (a−4)(x^2 +x+1)^2 =0 has at least one root , then find complete set of values of a.

If:(x2+x+2)2−(a−3)(x2+x+1)(x2+x+2)+(a−4)(x2+x+1)2=0hasatleastoneroot,thenfindcompletesetofvaluesofa.

Commented by rahul 19 last updated on 16/Mar/18

my try : let x^2 +x+1= t. ⇒ (t+1)^2 −(a−3)t(t+1)+(a−4)t^2 =0 since it has atleast one root, ⇒ Discriminant =0 ⇒(a−3)^2 t^2 −4t^2 (a−4)≥0 ⇒a^2 t^2 +25t^2 −10at^2 ≥0 ⇒t^2 (a−5)^2 ≥0 ⇒a≥5 ( t≠0) hence range = [5,∞). but correct ans. is (5,((19)/3)].

mytry:letx2+x+1=t.⇒(t+1)2−(a−3)t(t+1)+(a−4)t2=0sinceithasatleastoneroot,⇒Discriminant=0⇒(a−3)2t2−4t2(a−4)⩾0⇒a2t2+25t2−10at2⩾0⇒t2(a−5)2⩾0⇒a⩾5(t≠0)hencerange=[5,∞).butcorrectans.is(5,193].

Commented by mrW2 last updated on 16/Mar/18

Discriminant may not contain the variable itself. it can only contain the constants. correct: ax^2 +bx+c=0 D=b^2 −4ac not correct: ax^2 +b(x+d)x+c=0 D=b^2 (x+d)^2 −4ac

Discriminantmaynotcontainthevariableitself.itcanonlycontaintheconstants.correct:ax2+bx+c=0D=b2−4acnotcorrect:ax2+b(x+d)x+c=0D=b2(x+d)2−4ac

Commented by rahul 19 last updated on 18/Mar/18

Okay sir.

Okaysir.

Answered by mrW2 last updated on 16/Mar/18

let t=x^2 +x+1 (t+1)^2 −(a−3)t(t+1)+(a−4)t^2 =0 t^2 +2t+1−(a−3)t^2 −(a−3)t+(a−4)t^2 =0 (5−a)t+1=0 (5−a)(x^2 +x+1)+1=0 (5−a)x^2 +(5−a)x+(6−a)=0 a≠5 D=(5−a)^2 −4(5−a)(6−a)≥0 (5−a)(3a−19)≥0 ⇒5<a≤((19)/3)

lett=x2+x+1(t+1)2−(a−3)t(t+1)+(a−4)t2=0t2+2t+1−(a−3)t2−(a−3)t+(a−4)t2=0(5−a)t+1=0(5−a)(x2+x+1)+1=0(5−a)x2+(5−a)x+(6−a)=0a≠5D=(5−a)2−4(5−a)(6−a)⩾0(5−a)(3a−19)⩾0⇒5<a⩽193

Commented by rahul 19 last updated on 18/Mar/18

thank u so much sir!

thankusomuchsir!

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