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# 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 Abhay Kumar
1 year, 9 months ago

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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 = $$\beta + n\beta = -\dfrac{b}{a} \Rightarrow \color{red}{\beta} = -\dfrac{b}{a(1 + n)}$$.
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.

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

$$\quad \Rightarrow \boxed{ b \geq 2\sqrt{ac}}$$ · 1 year, 9 months ago

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 = $$2c - a - b = 0$$.
$$\quad \quad \quad \quad \Rightarrow \color{blue}{c} = \dfrac{a + b}{2}$$.

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

$$\quad \quad \quad \quad \quad \quad= -\dfrac{1}{2}(a^2 + b^2)$$ · 1 year, 9 months ago

Thank u very much. :) · 1 year, 9 months ago

For the second one, you get the root as 1 by observation and then just apply Vieta's and get the answer · 1 year, 9 months ago

Can u please show it. · 1 year, 9 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 · 1 year, 9 months ago