Question

Find the zeros of the quadratic polynomial f(x)=x2-3x-28 and verify the relationship between the zeros and the coefficients of the polynomial.

Answer 1

Answer:

f(x) = x² - 3x - 28

To find zero

put f(x) = 0

x² - 3x - 28 = 0

x² - 7x + 4x - 28 = 0

x ( x - 7 ) + 4 (x  - 7 ) = 0

( x - 7 )( x + 4 ) = 0

x = 7 & -4

let α = 7 and β = -4

Now to verify the relation

sum of zeroes = α + β = 7 + (-4) = 3

$\frac{-coefficient\:of\:x}{Coefficient\:of\:x^2}=\frac{-(-3)}{1}=3$

Product of zeroes = α . β = 7 × (-4) = -28

$\frac{constant\:term}{Coefficient\:of\:x^2}=\frac{-28}{1}=-28$

sum of zeroes =  $\frac{-coefficient\:of\:x}{Coefficient\:of\:x^2}$

Product of zeroes =  $\frac{constant\:term}{Coefficient\:of\:x^2}$

Hence proved

Answer 2

Step-by-step explanation:

$Given \: polynomial:\\f(x)=x^{2}-3x-28\\=x^{2}-7x+4x-28\\=x(x-7)+4(x-7)\\=(x-7)(x+4)$

$So,the\:value \:of\:f(x)\:is\\zero\:when \:x-7=0\:Or\:x+4=0$

$i.e., when \:x=7 \:Or \:x = -4$

Therefore,.

$the \:zeroes \:of \:f(x)\:are\\7\:and\:-4$

$Now,\\sum\:of\:the\:zeroes \\=7+(-4)\\=7-4\\=3\\=\frac{-(Coefficient\:of\:x)}{coefficient\:of\:x^{2}}$

$Product\:of\: zeroes\\=7\times (-4)\\=-28\\=\frac{Constant\:term}{coefficient\:of\:x^{2}}$

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