Question

3. If one of the zeroes of quadratic polynomial x2 + (a + 1)x -6 is 2, then what is the value of a?

please answer it

Answer 1

Given :

• x²+ (a +1 )x -6.
• one zero of given polynomial be 2.

To find :

The value of a.

Solution :

We have equation,

$\tt\longmapsto {x}^{2} + (a + 1)x - 6.$

When, We putting the value 2 at respect to x, then our Equation become 0.

$\tt\longmapsto {2}^{2} + (a + 1)(2) - 6= 0.$

$\tt\longmapsto 4 + 2a + 2 - 6 = 0.$

$\tt\longmapsto 0 + 2a = 0.$

$\tt\longmapsto a = 0.$

Therefore, the value of a become 0.

Answer 2

Given :

If one of the zeroes of quadratic polynomial x2 + (a + 1)x -6 is 2, then what is the value of a?

To find :

Find the value of a

Solution :

Given equation → x² + (a + 1)x - 6

Putting zero in given equation

=> x² + (a + 1)x - 6 = 0

=> (2)² + (a +1)×2 - 6 = 0

=> 4 + 2a + 2 - 6 = 0

=> 4 + a - 4 = 0

=> a = 0

Hence, required value of a = 0