Question

find the zeroes of (
$b {}^{2} - 17$
) and verify the relationship between the zeroes and the coefficients​

Answer 1

Here is ur answer

${b}^{2} - 17 \\ = (b - \sqrt{17} ) \: (b + \sqrt{17} ) \\ for \: zeros \: \\ b = \sqrt{17} and \: second \: zero \: b \: = - \sqrt{17} \\ therefore \: \alpha = \sqrt{17} and \: \beta = \sqrt{17} \\ verifying \: relationship \\ = > \: \alpha + \beta = \frac{ - b}{a} \\ \sqrt{17} + ( - \sqrt{17} ) = 0 \\ = > \alpha \times \beta = \frac{c}{a} \\ \sqrt{17} X( - \sqrt{17} )= - 17 \\ now \: lhs \: = rhs \: in \: both \: the \: cases \\ hence \: verified$

➰hope it helps u ❤➰

Answer 2

Let p(b) be the given polynomial.

Let @ and ß be the zeros of given polynomial.

p(b)=b^2 -17

= (b)^2-(√17)^2

=(b+√17)(b-√17)

We know that,

→Every polynomial expression is equal to zero.

Implies,

p(b)=0

=>(b+√17)(b-√17)=0

=>b+√17=0 or, b-√17=0

=>b= -√17 and, b=√17

or,b=±17

=>@=√17 and, ß= -√17

Verification:

Sum of zeros:

@ + ß

=√17+(-√17)

=0

Product of zeros:

@ß=(√17)(-√17)= -17