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

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

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

**4.**If \(Sinx + Siny=a\) and \(Cosx + Cosy=b\),Then find the roots of equation \((a^2+b^2+2b)t^2-4at+a^2+b^2-2b=0\)

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## Comments

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TopNewestA - 3. Let the one root of given equation is \(\beta\), then the other root must be n\(\beta\).

Using Vieta's formula, Sum of roots = \(\beta + n\beta = -\dfrac{b}{a} \Rightarrow \color{red}{\beta} = -\dfrac{b}{a(1 + n)}\).

Product of roots = \(n\color{red}{\beta}^2 = \dfrac{c}{a}\).

\( \Rightarrow n\left(\dfrac{-b}{a(1 + n)}\right)^2 = \dfrac{c}{a}\).

\( \Rightarrow nb^2 = ac + acn^2 + 2anc\).

\( \Rightarrow acn^2 + (2ac - b^2)n + ac = 0\).

For real value of n, discriminant \(\geq\) 0.

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

\(\quad \Rightarrow \boxed{ b \geq 2\sqrt{ac}}\)

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Simplified expression is,

\( \Rightarrow x^2 + x(a + b - 2c) + (ab - ac - bc) = 0 \).

Using vieta's formula,

Sum of roots = \( 2c - a - b = 0\).

\(\quad\)Product of roots = \( ab - a\color{blue}{c} - b\color{blue}{c}\).

\( \quad \quad \quad \quad \quad \quad= ab - a\left( \dfrac{a+b}{2}\right) - b\left(\dfrac{a+b}{2}\right) \)

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

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Thank u very much. :)

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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

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Can u please show it.

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

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