Help me.

1.The sum of the roots of the equation 1x+a+1x+b=1c\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c} is zero.Prove that the product of the roots is 12(a2+b2)-\frac{1}{2}(a^2+b^2).

2.If the roots of the equation p(qr)x2+q(rp)x+r(pq)=0p(q-r)x^2+q(r-p)x+r(p-q)=0 be equal , then show that 1p+1r=2q\frac{1}{p}+\frac{1}{r}=\frac{2}{q}.

3.Find the condition that one root of ax2+bx+c=0ax^2+bx+c=0 shall be n'n' times the other.

4.If Sinx+Siny=aSinx + Siny=a and Cosx+Cosy=bCosx + Cosy=b,Then find the roots of equation (a2+b2+2b)t24at+a2+b22b=0(a^2+b^2+2b)t^2-4at+a^2+b^2-2b=0

Note by Abhay Kumar
4 years ago

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  1. First, simplify the given expression by taking L.C.M and then cross multiplication,
    Simplified expression is,
    x2+x(a+b2c)+(abacbc)=0 \Rightarrow x^2 + x(a + b - 2c) + (ab - ac - bc) = 0 .
    Using vieta's formula,
    Sum of roots = 2cab=0 2c - a - b = 0.
c=a+b2\quad \quad \quad \quad \Rightarrow \color{#3D99F6}{c} = \dfrac{a + b}{2}.

\quadProduct of roots = abacbc ab - a\color{#3D99F6}{c} - b\color{#3D99F6}{c}.
=aba(a+b2)b(a+b2) \quad \quad \quad \quad \quad \quad= ab - a\left( \dfrac{a+b}{2}\right) - b\left(\dfrac{a+b}{2}\right)

=12(a2+b2) \quad \quad \quad \quad \quad \quad= -\dfrac{1}{2}(a^2 + b^2)

Akhil Bansal - 4 years ago

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Thank u very much. :)

Abhay Kumar - 4 years ago

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A - 3. Let the one root of given equation is β\beta, then the other root must be nβ\beta.
Using Vieta's formula, Sum of roots = β+nβ=baβ=ba(1+n)\beta + n\beta = -\dfrac{b}{a} \Rightarrow \color{#D61F06}{\beta} = -\dfrac{b}{a(1 + n)}.
Product of roots = nβ2=can\color{#D61F06}{\beta}^2 = \dfrac{c}{a}.
n(ba(1+n))2=ca \Rightarrow n\left(\dfrac{-b}{a(1 + n)}\right)^2 = \dfrac{c}{a}.
nb2=ac+acn2+2anc \Rightarrow nb^2 = ac + acn^2 + 2anc.
acn2+(2acb2)n+ac=0 \Rightarrow acn^2 + (2ac - b^2)n + ac = 0.

For real value of n, discriminant \geq 0.

(2acb2)24(ac)20 \Rightarrow (2ac - b^2)^2 - 4(ac)^2 \geq 0.

b2ac\quad \Rightarrow \boxed{ b \geq 2\sqrt{ac}}

Akhil Bansal - 4 years ago

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For the second one, you get the root as 1 by observation and then just apply Vieta's and get the answer

Harshit Singhania - 4 years ago

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Can u please show it.

Abhay Kumar - 4 years ago

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Sum of roots=2=pqrqpqrp2pq2rp=pqrqpq+rq=2rp2= \frac {pq-rq}{pq-rp} 》》2pq-2rp=pq-rq》》pq+rq=2rp Divide by pqr to get required result

Harshit Singhania - 4 years ago

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