Question

verify the relation between zeros and coefficient ​

Answer 1

Answer:

Consider quadratic polynomial

P(x) = 2x2 – 16x + 30.

Now, 2x2 – 16x + 30 = (2x – 6) (x – 3)

= 2 (x – 3) (x – 5)

The zeros of P(x) are 3 and 5.

Sum of the zeros = 3 + 5 = 8 = \frac { -\left( -16 \right) }{ 2 } = \text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\

Product of the zeros = 3 × 5 = 15 = \frac { 30 }{ 2 } = \left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\

So if ax2 + bx + c, a ≠ 0 is a quadratic polynomial and α, β are two zeros of polynomial then

\alpha +\beta =-\frac { b }{ a }

\alpha \beta =\frac { c }{ a }

In general, it can be proved that if α, β, γ are the zeros of a cubic polynomial ax3 + bx2 + cx + d, then

\alpha +\beta +\gamma =\frac { -b }{ a }

\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a }

\alpha \beta \gamma =\frac { -d }{ a }

Note: \frac { b }{ a } , \frac { c }{ a } and \frac { d }{ a } are meaningful because a ≠ 0.

Answer 2

Answer:

just check -b/a=P+Q  c/a=PQ if P,Q are zeroes and a,b,c are coff.

Step-by-step explanation: