Question

find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients of the polynomials



1. p(x) = 8x square- 19x - 15
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Answer 1

Step-by-step explanation:

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Answer 2

S O L U T I O N :

We have p(x) = 8x² - 19x - 15

Zero of the polynomial p(x) = 0

Using quadratic formula :

Compared with ax² + bx + c

• a = 8
• b = -19
• c = -15

Now;

$\longrightarrow\sf{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }\\\\\\\longrightarrow\sf{x=\dfrac{-(-19)\pm\sqrt{(-19)^{2}-4\times 8\times (-15) } }{2\times 8} }\\\\\\\longrightarrow\sf{x=\dfrac{19\pm\sqrt{361+480} }{16} }\\\\\\\longrightarrow\sf{x=\dfrac{19\pm\sqrt{841} }{16} }\\\\\\\longrightarrow\sf{x=\dfrac{19\pm29}{16} }\\\\\\\longrightarrow\sf{x=\dfrac{19+29}{16} \:\:Or\:\:x=\dfrac{19-29}{16} }\\\\\\\longrightarrow\sf{x=\cancel{\dfrac{48}{16}} \:\:Or\:\:x=\cancel{\dfrac{-10}{16}} }$

$\longrightarrow\bf{x=3\:\:\:Or\:\:\:-5/8}$

∴ The α = 3 and β = -5/8 are the zeroes of the polynomial.

As the given quadratic polynomial as we compared with ax² + bx + c;

• a = 8
• b = -19
• c = -15

So;

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

$\mapsto\rm{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{3+\bigg(-\frac{5}{8} \bigg)=\dfrac{-(-19)}{8} }\\\\\\\mapsto\rm{3-\dfrac{5}{8} =\dfrac{19}{8} }\\\\\\\mapsto\rm{\dfrac{24-5}{8} =\dfrac{19}{8} }\\\\\\\mapsto\bf{\dfrac{19}{8} =\dfrac{19}{8}}$

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

$\mapsto\rm{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{3\times \bigg(-\dfrac{5}{8} \bigg)=\dfrac{-15}{8} }\\\\\\ \mapsto\bf{\dfrac{-15}{8} =\dfrac{-15}{8}}$

Thus;

Relationship between zeroes and coefficient is verified .