Question

Find the zeros of the polynomial x2 +
coefficients and zeros of the polynomial.
x -2, and verify the relation between the
​

Answer 1

AnsWer :

x = -2 and x = 1.

Concepts Required :

$\tt \dagger \: \: \: \: \: sum \: of \: zeros.$

$\tt\alpha + \beta = \dfrac{ - b}{a} = \dfrac{coefficient \: of \: {x}^{2} }{coefficient \: of \: x}$

$\tt\dagger \: \: \: \: \: product\: of \: zeros.$

$\tt\alpha \beta = \dfrac{ c}{a} = \dfrac{contant \: term }{coefficient \: of \: x}$

Solution :

We have,

$\tt \dagger \: \: \: \: \: {x}^{2} + x - 2 = 0.$

Let's try to Find there Zeros,

$\tt \longmapsto {x}^{2} + x - 2 = 0.$

$\tt \longmapsto {x}^{2} + 2x - x - 2 = 0.$

$\tt \longmapsto x(x + 2) - 1(x + 2) = 0.$

$\tt \longmapsto (x - 1)(x + 2) = 0.$

$\rule{90}1$

Either,

$\tt\mapsto x - 1= 0.$

$\tt \mapsto x = 1.$

Or,

$\tt \mapsto x + 2 = 0.$

$\tt \mapsto x = - 2.$

Let,

$\dagger \: \: \: \: \: \: \: \: \alpha = - 2. \: \: \: \: \: and \: \: \: \: \: \: \beta = 1.$

$\rule{90}1$

Sum of Zeros,

$\tt \longmapsto \alpha + \beta = \dfrac{ - b}{a}$

$\tt \longmapsto - 2 + 1 = 1.$

$\tt \longmapsto 1 = 1.$

$\rule{90}1$

Product of Zeros.

$\tt\longmapsto \alpha \beta = \dfrac{c}{a}$

$\tt \longmapsto - 2 \times 1 = - 2.$

$\tt \longmapsto - 2 = - 2.$

Hence Verify.