Question

find the zeroes of the quadratic polynomial
$2 {x}^{2} - 7x + 5$
and verify the relationship between zeroes and coefficients​

Answer 1

Step-by-step explanation:

2x^2-7x+5=0

2x^2-2x-5x+5=0

2x(x-1)-5(x-1)=0

(2x-5)(x-1)=0

x=(5/2) or x=1

Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

The quadratic polynomial is 2x² - 7x + 5 and verify the relationship between zeroes and coefficient.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The zeroes of the polynomial.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

We have p(x) = 2x² - 7x + 5

Zero of the polynomial is p(x) = 0

So;

$\leadsto\sf{2x^{2} -7x+5=0}\\\\\leadsto\sf{2x^{2} -2x-5x+5=0}\\\\\leadsto\sf{2x(x-1)-5(x-1)=0}\\\\\leadsto\sf{(x-1)(2x-5)=0}\\\\\leadsto\sf{x-1=0\:\:Or\:\:2x-5=0}\\\\\leadsto\sf{x=1\:\:\:Or\:\:\:2x=5}\\\\\leadsto\sf{\pink{x=1\:\:\:Or\:\:\:x=\dfrac{5}{2} }}$

∴ The α = 1 and β = 5/2 are the zeroes of the polynomial.

Now;

As the given quadratic polynomial as we compared with ax² + bx + c = 0

• a = 2
• b = -7
• c = 5

$\star\:\bf{\orange{\large{\underline{\mathcal{SUM\:OF\:THE\:ZEREOS\::}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x^{2} }{Coefficient\:of\:x} }\\\\\\\mapsto\sf{1+\dfrac{5}{2} =\dfrac{-(-7)}{2} }\\\\\\\mapsto\sf{\dfrac{2+5}{2} =\dfrac{7}{2} }\\\\\\\mapsto\sf{\pink{\dfrac{7}{2} =\dfrac{7}{2} }}$

$\star\:\bf{\orange{\large{\underline{\mathcal{PRODUCT\:OF\:THE\:ZEREOS\::}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:x} }\\\\\\\mapsto\sf{1\times \dfrac{5}{2} =\dfrac{5}{2} }\\\\\\\mapsto\sf{\pink{\dfrac{5}{2} =\dfrac{5}{2}}}$

Thus;

Relationship between zeroes and coefficient is verified.