Question

find the xeros of each of the following polynomials and verify the relationsp between the xeros and the coefficients if the polynomial 2x2-9x-45

Answer 1

Given equation :- 2x² - 9x - 45

• Factorising by Middle term splitting


2x² - 9x - 45 = 0

2x² - 15x + 6x - 45 = 0

x ( 2x - 15 ) + 3 ( 2x - 15 ) = 0

( x + 3 ) ( 2x - 15 ) = 0


# ( x + 3 ) = 0
x = - 3

# ( 2x - 15 ) = 0

x = 15/2


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• Relation between it's coefficient :-


✴
$sum \: of \: zeros = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$


° Taking LHS :-

Sum of Zeros = - 3 + 15/2

= ( - 6 + 15 )/2

= 9/2

° Taking RHS :-


$\frac{ - coeff \: \: of \: x}{coeff \: of \: {x}^{2} }$

= 9/2



LHS = RHS


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


° Taking LHS :-

Product of Zeros = - 3 × 15/2

= - 45/2


° Taking RHS :-

$\frac{constant \: term}{coeff \: of \: {x}^{2} }$

= - 45/2

LHS = RHS


Since , LHS = RHS in both case

Hence verified the relation between the zeros and the coefficient of the equation !!!

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

f(x) = 2x² - 9x - 45

$\fbox{Splitting\;Middle\;Term}$

f(x) = 2x² - 15x + 6x - 45

f(x) = x(2x - 15) + 3(2x - 15)

f(x) = (x + 3)(2x - 15)

★Zeroes are :-

$\tt{\rightarrow -3\;and\;\dfrac{15}{2}}$

$\Large{\fbox{Relationship}}$

★Sum of zeroes = α + β

$\tt{\rightarrow\alpha+\beta=-\dfrac{b}{a}}$

$\tt{\rightarrow (-3)+\dfrac{15}{2}=\dfrac{-(-9)}{2}}$

$\tt{\rightarrow\dfrac{9}{2}=\dfrac{9}{2}}$

$\fbox{Proved}$

★Product of zeroes = α × β

$\tt{\rightarrow\alpha\times\beta=\dfrac{c}{a}}$

$\tt{\rightarrow (-3)\times\dfrac{15}{2}=\dfrac{-45}{2}}$

$\tt{\rightarrow -\dfrac{45}{2}=\dfrac{-45}{2}}$

$\fbox{Proved}$

$\Large{\fbox{Hence\; Verified}}$