Question

findfind the zero of the quadratic polynomial 5 x square - 8 x minus 4 and verify the relationship between the zeros and their coefficients
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Answer 1

S O L U T I O N :

We have quadratic polynomial p(x) = 5x² - 8x - 4 & zero of the polynomial p(x) = 0 .

$\underline{\underline{\tt{Using\:\:by\:\:factorisation\:\:method\::}}}$

$\longrightarrow\sf{5x^{2} - 8x - 4 = 0}$

$\longrightarrow\sf{5x^{2} -10x + 2x - 4 = 0}$

$\longrightarrow\sf{5x(x - 2) + 2(x-2) = 0}$

$\longrightarrow\sf{(x-2) (5x + 2) = 0}$

$\longrightarrow\sf{x-2=0\:\:\:Or\:\:\:5x + 2 = 0}$

$\longrightarrow\sf{x=2\:\:\:Or\:\:\:5x =-2}$

$\longrightarrow\bf{x=2\:\:\:Or\:\:\:x =-2/5}$

∴ α = 2 & β = -2/5 are the two zeroes of the given polynomial .

As we know that given polynomial compared with ax² + bx + c;

• a = 5
• b = -8
• c = -4

Now,

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta =\dfrac{-b}{a} = \bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{2+ \bigg(-\dfrac{2}{5} \bigg) =\dfrac{-(-8)}{5} }$

$\mapsto\tt{2-\dfrac{2}{5} =\dfrac{8}{5} }$

$\mapsto\tt{\dfrac{10-2}{5} =\dfrac{8}{5} }$

$\mapsto\bf{\dfrac{8}{5} =\dfrac{8}{5} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta =\dfrac{c}{a} = \bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{2\times \bigg(-\dfrac{2}{5} \bigg) =\dfrac{-4}{5} }$

$\mapsto\bf{\dfrac{-4}{5} =\dfrac{-4}{5} }$

Thus;

The relationship between zeroes & coefficient are verified .