Question

Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
(iv) $6x^{2}-3-7x$
(v) $p(x)=x^{2} +2\sqrt{2x}-6$
(vi) $q(x)=\sqrt{3x^{2}}+10x+ 7\sqrt{3}$

Answer 1

SOLUTION :  

(iv) Given : f(x) =  6x² − 3− 7x

f(x) = 6x² −7x− 3

f(x)= 6x² - 9x + 2x -3

f(x)= 3x ( 2x - 3) + 1 (2x - 3)

f(x) = (3x + 1)(2x– 3)

To find zeroes,  put f(x) = 0

(3x + 1)(2x– 3) = 0  

(3x + 1) = 0   or   (2x– 3) = 0

3x = -1  or  2x = 3

x = -1/3  or x = 3/2

Hence, Zeroes of the polynomials are α = - ⅓  and  β = 3/2

VERIFICATION :  

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β = −coefficient of x / coefficient of x²

−1/3 + 3/2 = −(−7)/6

(-2+ 9)/6 = 7/6

7/6 = 7/6

Product of the zeroes = constant term/ Coefficient of x²

α β = constant term / Coefficient of x²

−1/3 × 3/2 = −3/6

−3/6 = −3/6

-½ = -½

Hence, the relationship between the Zeroes and  its coefficients is verified.

(v)  GIVEN : p(x)= x² + 2√2x - 6

= x² + 3√2x + √2x - 6

= x(x + 3√2)–√2(x + 3√2)

= (x + 3√2)(x–√2)

To find zeroes,  put p(x) = 0

(x + 3√2)(x–√2) = 0

(x + 3√2)  = 0   or   (x–√2) = 0

x = -3√2   or   x = √2

Hence, Zeroes of the polynomials are α = -3√2  and  β = √2

VERIFICATION :  

Sum of the zeroes = − coefficient of x / coefficient of x²

α + β = −coefficient of x / coefficient of x²

−3√2 + √2 = −2√2/1

√2 (-3 +1) = −2√2/1

√2 × -  2 = −2√2/1

-2√2 = -2√2

Product of the zeroes = constant term/ Coefficient of x²

α β = constant term / Coefficient of x²

−3√2 × √2 = −6/1

-3× 2 = -6

-6 = -6

Hence, the relationship between the Zeroes and  its coefficients is verified.

 

(vi) Given : q(x) = √3x² + 10x + 7√3

= √3x² + 3x + 7x + 7√3

= √3x(x + √3 ) + 7 ( x + √3)

 [√3 ×√3 = 3]

= (x + √3) (√3x +7)

To find zeroes,  put q(x) = 0

(x + √3) (√3x +7) = 0

(x + √3)   = 0   or   (√3x +7) = 0

x = - √3  or  x = -7/√3

Hence, Zeroes of the polynomials are α = - √3  and  β = -7/√3

VERIFICATION :  

Sum of the zeroes = − coefficient of x / coefficient of x²

α + β = −coefficient of x / coefficient of x²

-√3 + -7/√3  = − (10) / √3

[(-√3 × √3 ) -7 ]/√3 =  − (10) / √3

(-3 - 7) /√3 =  − 10 / √3

-10/√3 =  − 10 / √3

Product of the zeroes = constant term/ Coefficient of x²

α β = constant term / Coefficient of x²

-√3 × -7/√3   = 7√3 /√3

7 = 7

Hence, the relationship between the Zeroes and  its coefficients is verified.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

hii dear


ur answer


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Given quadratic equation = √3x² - 8x + 4√3 = 0
We should factorize the equation first.
          =√3x²-6x-2x+4√3 = 0
          = √3x(x-2√3)-2(x-2√3) = 0 
          = (√3x-2) (x-2√3) = 0
          = (√3x-2) = 0 , (x-2√3) = 0
          = x = 2/√3 , x = 2√3.
⤴⤴⤴⤴⤴⤴⤴⤴⤴⤴⤴⤴⤴⤴⤴