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`        Find all pairs (a,b) of real nos. such that whatever k is a root of x2+ax+b=0 k2-2 is also a root of the equation`
10 years ago

```							Let one root be k. Now the other root has to be equal to k^2 - 2. This is because a quadratic equation can have only two roots(equal or unequal). Now using the standard results, we get k+k^2-2 = -a (Sum of the roots = -(b/a). Here b=a, a=1.) Similarly, k(k^2 - 2)=b. These two equations now form an identity in k. For any value of k, we get a definite value for a and b. Thus there are infinitely many possibilities for the values of a and b.

As a trial, try it out with k=3. we get a=-10, b=21. So, the quadratic equation formed is

x^2-10x+21=0. Factorising, we get (x-3)(x-7)=0. So x=3 or x=7. Now, we had taken k=3. So k^2-2 will be 7 which is also the root of the equation. Thus, the relation that we have found for a and b holds.
```
10 years ago
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