# Prove The Following (Quadratic Equations)

This note facilitates you to prove good questions based on Quadratic Equations. So please use this note. I will try my best to post these types of notes from now.

$$\color{green}\text{Prove the Following :}$$

1. If the sum of the roots of the equation $$px^2 + qx + r = 0$$ be equal to the sum of their squares, show that $$2pr = pq + q^2$$

2. If the difference of roots of $$x^2 - bx + c = 0$$ be same as that of the roots of $$x^2 - cx + b = 0$$ then prove that $$b + c = -4$$ and $$b = c$$

3. If the roots of the equation $$\dfrac{1}{(x + p)} + \dfrac{1}{(x + q)} = \dfrac{1}{r}$$ are equal in magnitude but opposite in sign, show that $$p + q = 2r$$ and product of the roots is equal to $$-\dfrac{1}{2}(p^2 + q^2)$$

4. If the ratio of the roots of the equation $$ax^2 + bx + c = 0$$ is $$r$$, then prove that $$\dfrac{(r + 1)^2}{r} = \dfrac{b^2}{ac}$$

5. If the ratio of the roots of the equation $$x^2 + px + q = 0$$ be equal to the ratio of roots of the equation $$x^2 + bx + c = 0$$, then prove that $$p^2c = b^2q$$

6. If the sum of the two roots of the equation $$x^3 - px^2 + qx - r = 0$$ is zero, then prove that $$pq = r$$

7. If the equations $$ax^2 + bx + c = 0$$ and $$bx^2 + cx + a = 0$$ have a common root than prove that $$a^3 + b^3 + c^3 = 3abc$$ and $$a = 0$$

If anybody proved any one of these I will mark that question as proved. But if you have another idea to the same question you can post it too. So I will also mention in how many methods the proof is proved. Try your best to prove these.

For more of these see my set Proof Based Notes

Note by Ram Mohith
3 months, 2 weeks ago

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## Comments

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Eliminate x and y sin(x) + sin(y) = a cos(x) + cos(y) = b tan(x) + tan(y) = c

- 3 months ago

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1) Let the roots of the equation $$px^2 + qx + r = 0$$ be $$a,b$$

Given : $$a + b = a^2 + b^2$$

$$\implies a + b = (a + b)^2 - 2ab$$

$$\implies -\dfrac{q}{p} = \dfrac{q^2}{p^2} - \dfrac{2r}{p}$$

$$\implies - pq = q^2 - 2pr$$ (Multiplying all terms by $$p^2$$)

$$\implies 2pr = pq + q^2$$

$$\color{green}\text{Hence Proved}$$

- 3 months, 1 week ago

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