Question

1. Find the ratio of sum and product of the zero of the
given polynomial 2√3x 7x² + √3​

Answer 1

Step-by-step explanation:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8

(ii) 4s2 – 4s + 1

(iii) 6x2 – 3 – 7x

(iv) 4u2 + 8u

(v) t2 – 15

(vi) 3x2 – x – 4

Sol. (i) x2 – 2x – 8

We have p(x) = x2 – 2x – 8

= x2 + 2x – 4x – 8 = x (x + 2) – 4 (x + 2)

= (x – 4) (x + 2)

For p(x) = 0, we have

(x – 4) (x + 2) = 0

Either x – 4 = 0 ⇒ x = 4

or x + 2 = 0 ⇒ x = – 2

∴ The zeroes of x2 2x – 8 are 4 and –2.

Now, sum of the zeroes

Thus, relationship between zeroes and the coefficients in x2 – 2x – 8 is verified.

(ii) 4s2 – 4s + 1

We have p(s) = 4s2 – 4s + 1

= 4s2 – 2s – 2s + 1 = 2S (2S – 1) –1 (2s – 1)

= (2s – 1) (2s – 1)

For p(s) = 0, we have,

Now,

Thus, the relationship between the zeroes and coefficients in the polynomial 4s2 – 4s + 1 is verified.

(iii) 6x2 – 3 – 7x

We have

p (x) = 6x2 – 3 – 7x

= 6x2 – 7x – 3

= 6x2 – 9x + 2x – 3

= 3x (2x – 3) + 1 (2x – 3)

= (3x + 1) (2x – 3)

For p (x) = 0, we have,

Answer 2

Answer:

-7/ $\sqrt{3}$

Step-by-step explanation:

Sum of zeroes = -7/(2 $\sqrt{3}$)

Product of zeroes =  $\sqrt{3}$ / 2

Their ratio = [-7/(2 $\sqrt{3}$)] / (1/2) = -14 / 2 $\sqrt{3}$ =  -7/ $\sqrt{3}$

So your answer is -7/ $\sqrt{3}$    ( rationalised form : (-7 $\sqrt{3}$  ) / 3)

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