# Help me.

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 $(a^2+b^2+2b)t^2-4at+a^2+b^2-2b=0$ Note by A Former Brilliant Member
5 years, 8 months ago

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

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

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

- 5 years, 8 months ago

Thank u very much. :)

- 5 years, 8 months ago

A - 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 = $\beta + n\beta = -\dfrac{b}{a} \Rightarrow \color{#D61F06}{\beta} = -\dfrac{b}{a(1 + n)}$.
Product of roots = $n\color{#D61F06}{\beta}^2 = \dfrac{c}{a}$.
$\Rightarrow n\left(\dfrac{-b}{a(1 + n)}\right)^2 = \dfrac{c}{a}$.
$\Rightarrow nb^2 = ac + acn^2 + 2anc$.
$\Rightarrow acn^2 + (2ac - b^2)n + ac = 0$.

For real value of n, discriminant $\geq$ 0.

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

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

- 5 years, 8 months ago

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

- 5 years, 8 months ago