Question

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​

Answer 1

Solution:-

Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.

->Let p(x) be a polynomial with any number of terms any number of degree.

(Now, zeroes of the polynomial will be the values of x at which p(x) = 0.If p(x) = ax2 + bx + c is a quadratic polynomial.)

roots are α and β

Sum of the roots = α + β = - b/aProduct of roots = αβ = c/a 

(i) x^2 - 4x + 2x - 8 = 0

( factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)

x(x - 4) + 2(x - 4) = 0= (x - 4)(x + 2)

 The value of x^2 - 2x - 8 is zero when x − 4 = 0 or x + 2 = 0, (x = 4 or x = −2)

Therefore, The zeroes of x2 - 2x - 8 are 4 and −2.

 Sum of zeroes = 4 + (-2) = 2 

$= \frac{ - ( - 2)}{1} = - \frac{( - coefficient \: of \: x)}{(coefficient \: of \: x {}^{2} )}$

Hence, it is verified that,

Sum of zeroes =

$- \frac{( \: coefficient \: of \: x)}{cofficient \: of \: x {}^{2}) }$

Product of zeroes =

$4 \times ( - 2) = - 8 = \frac{( - 8)}{1}$

Hence,

$product \: of \: zeros \: = \: \frac{constant \: term}{coefficient \: of \: x {}^{2} }$

(ii) 4s2 - 4s + 1

= (2s)^2 - 2(2s)1 + 1^2

As, we know (a - b)^2 = a^2 - 2ab + b^2, the above equation can be written as:-

= (2s - 1)^2

The value of 4s^2 − 4s + 1 is zero when 2s − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4s^2 − 4s + 1 are 

Sum of zeroes

$\frac{1}{2} + \frac{1}{2} = 1 = - \frac{( - 4)}{4} = \frac{(coefficient \: of \: s}{coefficient \: of \: s {}^{2} }$

Peoduct of zeroes

$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{(constant \: term)}{coefficient \: of \: s {}^{2} }$

Hence, verified!

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Note:- 3rd , 4th ,5th , 6th parts are shown in the above attachment.

Answer 2

Step-by-step explanation:

heey buddy

hope it's useful