Question

verify whether 2,3,1/2 are tje zeros of the polynomial 2x³-11x²+17x-6 and verify the relationship between zeros and coefficients​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

2,3, 1/2 are the zeroes of the polynomial 2x³ - 11x² + 17x - 6 and verify the relationship between zeroes and coefficient.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

Verify.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have p(x) = 2x³ - 11x² + 17x - 6

Putting the given zero in above equation :

p(2) = 0

$\longrightarrow\sf{2(2)^{3} -11(2)^{2} +17(2)-6=0}\\\\\longrightarrow\sf{2\times 8-11\times 4+34-6=0}\\\\\longrightarrow\sf{16-44+28=0}\\\\\longrightarrow\sf{44-44=0}\\\\\longrightarrow\sf{0=0}$

p(3) = 0

$\longrightarrow\sf{2(3)^{3} -11(3)^{2} +17(3)-6=0}\\\\\longrightarrow\sf{2\times 27-11\times 9+51-6=0}\\\\\longrightarrow\sf{54-99+45=0}\\\\\longrightarrow\sf{99-99=0}\\\\\longrightarrow\sf{0=0}$

p(1/2) = 0

$\longrightarrow\sf{2\bigg(\dfrac{1}{2} \bigg)^{3} -11\bigg(\dfrac{1}{2} \bigg)^{2} +17\bigg(\dfrac{1}{2} \bigg)-6=0}\\\\\\\longrightarrow\sf{\cancel{2}\times \dfrac{1}{\cancel{8}} -11\times \dfrac{1}{4} +\dfrac{17}{2} -6=0}\\\\\\\longrightarrow\sf{\dfrac{1}{4} -\dfrac{11}{4} +\dfrac{17}{2} -6=0}\\\\\\\longrightarrow\sf{\dfrac{1-11+34-24}{4} =0}\\\\\\\longrightarrow\sf{\dfrac{35-35}{4} =0}\\\\\\\longrightarrow\sf{\dfrac{0}{4} =0}\\\\\\\longrightarrow\sf{0=0}$

Now;

We know that the cubic polynomial as compared with ax³ + bx² + cx + d;

• a = 2
• b = -11
• c = 17
• d = -6

$\dag\:\underline{\underline{\bf{\green{Sum\:of\:the\:zeroes\::}}}}$

$\implies\sf{\alpha +\beta +\gamma=\dfrac{-b}{a}=\dfrac{(-Coefficient\:of\:(x)^{2}) }{Coefficient\:of\:(x)^{3}} }\\\\\\\implies\sf{2+3+\dfrac{1}{2} =\dfrac{-(-11)}{2} }\\\\\\\implies\sf{5+\dfrac{1}{2} =\dfrac{11}{2} }\\\\\\\implies\sf{\dfrac{10+1}{2} =\dfrac{11}{2} }\\\\\\\implies\sf{\pink{\dfrac{11}{2} =\dfrac{11}{2} }}$

$\dag\:\underline{\underline{\bf{\green{Sum\:of\:product\:of\:zeroes\::}}}}$

$\implies\sf{\alpha \beta +\beta \gamma+\alpha \gamma=\dfrac{c}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:(x)^{3} } }\\\\\\\implies\sf{(2)(3)+(3)\bigg(\dfrac{1}{2} \bigg)+2\bigg(\dfrac{1}{2} \bigg)=\dfrac{17}{2} }\\\\\\\implies\sf{6+\dfrac{3}{2} +\cancel{\dfrac{2}{2}} =\dfrac{17}{2} }\\\\\\\implies\sf{6+\dfrac{3}{2} +1=\dfrac{17}{2} }\\\\\\\implies\sf{\dfrac{12+3+2}{2} =\dfrac{17}{2} }\\\\\\\implies\sf{\pink{\dfrac{17}{2} =\dfrac{17}{2} }}$

$\dag\:\underline{\underline{\bf{\green{Product\:of\:zeroes\::}}}}$

$\implies\sf{\alpha \beta \gamma=\dfrac{-d}{a} =\dfrac{(-Constant\:term)}{Coefficient\:of\:(x)^{3} } }\\\\\\\implies\sf{2\times 3\times \dfrac{1}{2} =\dfrac{-(-6)}{2} }\\\\\\\implies\sf{\pink{\dfrac{6}{2} =\dfrac{6}{2} }}$

Thus;

Relationship between zeroes and coefficient is verified .

Answer 2

Given:

2,3,1/2 are tje zeros of the polynomial 2x³-11x²+17x-6 and verify the relationship between zeros and coefficients.

To find:

verify

Explanation:

we have p(x) = 2x³ - 11x² + 17x - 6

putting the given zero in above equation.

p(2) =0

2(2)³ - 11(2)² + 17(2) - 6

2 × 8 - 11 × 4 + 34 - 6 =0

16 - 44 + 28 = 0

44 - 44 = 0

0 = 0

p(3) =0

2(3)³ - 11(3)² + 17(3) - 6

2 × 27 - 11 × 9 + 51 - 6 =0

54 - 99 + 45 = 0

99 - 99 = 0

0 = 0

p(1/2) =0

2(1/2)³ - 11(1/2)² + 17(1/2) - 6

2 × 1/8 - 11 × 1/4 + 17/2 - 6 =0

1/4 - 11/4 + 17/2 = 0

1 - 11 + 34 - 24 /4 = 0

35 - 35 /4 =0

0/4 = 0

0= 0

Now,we know that the cubic polynomial as compared with ax³ + bx² + CX + d;

• a = 2

• b = -11

• c = 17

• d = -6

Sum of the zeroes;

$\alpha + \beta + \gamma = - b \div a = ( - coefficient \: of \: (x)^{2} ) \div (coefficient \: of \: x ^{3} )$

2 + 3 + 1/2 = -(-11)/2

5 + 1/2 = 11/2

10 + 1 /2 = 11/2

11/2 = 11/2

$\alpha \beta + \beta \gamma + \alpha \gamma = c \div a = coefficient \: of \: x \: \div coefficient \: of \: x ^{3}$

(2)(3) + (3)(1/2) + 2(1/2) = 17/2

6 + 3/2 + 2/2 = 17 /2

6 + 3/2 + 1 = 17/2

12 + 3 + 2 /2 = 17/2

17/2 = 17/2

product of zeroes;

$\alpha \beta \gamma = - d \div a \: - coefficient term \: \div coefficient \: of \: x ^{3}$

2 × 3 × 1/2 = -(-6)/2

6/2 = 6/2

Thus,

Relationship between zeroes and coefficient is verified.

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