Question

13. If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and -3, then find the value
of a and b.

Answer 1

The given polynomial,

${x}^{2} + (a + 1)x + b = 0$

We have the given roots as 2 and -3.

Which means that if you were to substitute these two values in the given polynomial for x you will get a 0 as result.

Let's do it one by one,

Taking x = 2

The polynomial becomes,

${2}^{2} + (a + 1)(2) + b = 0 \\ \\ 4 + 2a + 2 + b = 0 \\ \\ 6 + 2a + b = 0 \\ \\ 2a + b = - 6$

Using x = -3

${( - 3)}^{2} + (a + 1)( - 3) + b = 0 \\ \\ 9 - 3a - 3 + b = 0 \\ \\ 6 - 3a + b = 0 \\ \\ - 3a + b = - 6$

In the end of substitution we end up with 2 equations.

So to get the value of a and b we must use them somehow,

If you try this your method may vary but I prefer elemination method,

So let's do it,

Subtracting 1st equation we got from 2nd,

$2a + b - ( - 3a + b) = - 6 - ( - 6) \\ \\ 2a + 3a = 0 \\ \\ 5a = 0 \\ \\ a = 0$

Using this value of a and the first equation we had gained,
$2(0) + b = - 6 \\ \\ b = - 6$

Therefore we get the final values as a = 0 and b = -6.



Hope I was able to make it easier for you.