Question

Give the correct relation between the zeroes and coefficients of a polynomial ax^(2)+bx+ 1.Match the APs given in column A with suitable common differences given in column B . qquad Column A Column B (A1) 2 -2 -6 -10 ... (B1) 2/3 (A2) a=-18 n=10 a_(n)=0 (B2)-5 (A3) a=0 a_(10)=6 (B3) 4 (A4) a_(2)=13 a_(4)=3 (B4)-4 (B5) 2 (B6) 1/2​

Answer 1

Answer:

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Answer 2

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.

Example 1: Find the zeros of the quadratic polynomial 6x2 – 13x + 6 and verify the relation between the zeros and its coefficients.

Sol. We have, 6x2 – 13x + 6 = 6×2 – 4x – 9x + 6

= 2x (3x – 2) –3 (3x – 2)

= (3x – 2) (2x – 3)

So, the value of 6x2 – 13x + 6 is 0, when

(3x – 2) = 0 or (2x – 3) = 0 i.e.,

When x = \frac { 2 }{ 3 } or \frac { 3 }{ 2 }

Therefore, the zeros of 6x2 – 13x + 6 are

\frac { 2 }{ 3 } and \frac { 3 }{ 2 }

Sum of the zeros

= \frac { 2 }{ 3 } + \frac { 3 }{ 2 } = \frac { 13 }{ 6 } = -\frac { \left( -13 \right) }{ 6 } = \text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\

Product of the zeros

= \frac { 2 }{ 3 } × \frac { 3 }{ 2 } = \frac { 6 }{ 6 } = \left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\

Step-by-step explanation:

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