Question

Find the zeros of the following quadratic polynomials and verify the relationship
between the zeros and the coefficients:

• x²+ 7x + 12
• x² - 2x -8



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Answer 1

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Step-by-step explanation:

The given quadratic polynomial is,

p(x) =4x² - 4x - 3

=> 4x² - 6x + 2x - 3

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

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

p(x) = 0

(2x + 1) = 0 or (2x-3) = 0

x=-1/2 or x= 3/2

Hence, -1/2 and 3/2 are the zeroes of p(x).

Sum of zeroes = -1/2 + 3/2 = 2/2 = 1 = Coefficient of x / Coefficient of x²

Product of zeroes = (-1/2)(3/2) = -3/4 = Constant term/ Coefficient of x²

Hope this would help you!!

Answer 2

Answer:

$\huge\boxed{\fcolorbox{red}{pink}{Your's Answer}}$

$\red {QUESTION}$

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:

• x²+ 7x + 12
• x² - 2x -8

$\green {ANSWER}$

• x²+ 7x + 12

Given,

x²+ 7x + 12

zeroes

x²+ 7x + 12= 0

(x+3)(x+4)= 0

x= -3,-4

verification,

sum of roots= $- \frac{b}{a} = - \frac{7}{1} = - 7$

product of roots= $\frac{c}{a} = \frac{12}{1} = 12$

HENCE PROVED:

• x² - 2x -8

Given,

x² - 2x -8

zeroes,

x² - 2x -8= 0

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

x= 4,-2

verification,

sum of roots= $- \frac{b}{a} = - \frac{( - 2)}{1} = 2$

product of roots= $\frac{c}{a} = \frac{ - 8}{1} = - 8$

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