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Question

# (a) a,b,c are lengths of sides of an scalene triangle. If equationx2+2(a+b+c)x+3λ(ab+bc+ac)=0has real and distinct roots, then the value of λ is given by :(a) λ<43(b) 113<λ<173(c) 116<λ<154(d) λ≥1(b) If a,b are the roots of x2−10cx−11d=0 and c,d are the roots of x2−10ax−11b=0, then find the value, of a+b+c+d. (a,b,c,d are distinct numbers).

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Solution

## A quadratic equation has real and distinct roots implies its Δ≥0.or 4(a+b+c)2−12λ(ab+bc+ca)≥0or (∑a2+2∑ab)−3λ∑ab≥0or ∑a2≥(3λ−2)∑ab or 3λ−2<∑a2∑ab.......(1)Now in any triangle sum of two sides is greater than the third.∴ a+b>c b+c>a, c+a>bor a>c−b, b>a−c, c>b−aor a2+b2+c2>2(a2+b2+c2)−2∑abor 2∑ab>∑a2 or ∑a2∑ab<2......(2)Hence from (1) and (2). 3λ−2<2 or λ<4/3.(b) Ans.1210Given a+b=10c, ab=11d.........(1) c+d=10a, cd=−11b........(2) ∴ a+b+c+d=10(a+c)......(3)So we have to find the value of a+c so that by (3) We get the value of a+b+c+d.From relations (1) and (2) on adding,we have b+d=9(a+c)........(4)and multiplying relations in (1) and (2) (ac)bd=121bd ⇒ ac=121.........(5)Multiplying (1) by a and (2) by c and adding, a2+c2+(ab+cd)=(10+10)acor a2+c2−11(b+d)=20ac by (1),(2),or (a+c)2−11×9(a+c)=22acor t2−99t−22(121)=0 by (4) and (5)or t2−(121−22)t−22(121)=0 (t−121)(t+22)=0 ∴ t=121=a+cHence from (3), a+b+c+d=10(a+c)=10×121=1210

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