If $a$ and $b$ are the zeroes of $x^2 + ax + b = 0$, then how many pairs of $(a,b)$ exist?
- One
- Two
- Three
- Infinitely many
Also, what are these pairs?
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Shaurya Gupta
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Just substitute $a$ and $b$ into the equation. You get two quadratic equations in $a$ and $b$ that are easy to solve. You should find 3 solution pairs. – Leen Droogendijk Nov 04 '13 at 13:55
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@LeenDroogendijk I found out two pairs:(0,0) and (1,-2). How do I find out the third one? – Shaurya Gupta Nov 04 '13 at 13:56
3 Answers
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Since $a, b$ are zeros, they each satisfy the equation:
$$x=a: \quad a^2 + a\cdot a + b = 0\iff 2a^2 + b = 0 \iff b = -2a^2$$
$$x = b:\quad b^2 + ab + b = 0$$
Substituting $b = -2a^2$ into the second equation: $$(-2a^2)^2 + a(-2a^2) + -2a^2 = 0 \iff 4a^4 -2a^3 - 2a^2 = 2a^2(2a^2 -a - 1) = 0$$
The right-hand factor itself factors nicely, giving you 3 solutions to $a$, and then you can find the corresponding value for $b$ by using $b = -2a^2$.
amWhy
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Sum of the roots is $a+b = -a \implies 2a+b = 0$.
Product of the roots is $ab = b \implies b = 0 \text{ or } a = 1$.
If $b=0$, then $2a+ 0 = 0 \implies a = 0$.
If $a=1$, then $2+b = 0 \implies b = -2$.
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So that means two pairs of (a,b) but the answer is three. I had already figured out these two but cant figure out the third one. – Shaurya Gupta Nov 04 '13 at 13:55
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1The third case, $a=b$, so they are both the same root of a quadratic equation. They are not the roots, as asked in the question. – Empy2 Nov 04 '13 at 14:18
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Hint: if $a$ and $b$ are the zeroes then $x^{2}+ax+b=\left(x-a\right)\left(x-b\right)$.
drhab
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