Question

Find the zeros of a quadratic polynomial 5x2 – 4 – 8x and verify the relationship between the
zeros and the coefficients of the polinomial.

Answer 1

Hey!

Given polynomial :- 5x² - 4 - 8x

Any Quadratic polynomial should be in the form of ax² + bx + c

So, the polynomial is 5x² - 8x - 4

Let's factorise it by middle term splitting method :)


5x² - 8x - 4

5x² - 10x + 2x - 4

5x ( x - 2 ) + 2 ( x - 2 )

( 5x + 2 ) ( x - 2 )

• ( 5x + 2 ) = 0

x = -2/5

• ( x - 2 ) = 0

x = 2



So, the zeros are -2/5 and 2 !!


° To verify :-

• Sum of Zeros =
$= \frac{ - coeff. \: \: of \: x}{coeff. \: of \: {x}^{2} }$


Taking LHS

° Sum of Zeros = -2/5 + 2

$= \frac{ - 2 + 10}{5} \\ \\ = \frac{8}{5}$

Now ,taking RHS

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


$= \frac{ - ( - 8)}{5} = \frac{8}{5}$


Hence , LHS = RHS !!



• Product of Zeros =
$\frac{constant \: term \: }{coeff. \: \: of \: {x}^{2} }$

Taking LHS

° Product of Zeros = -2/5 × 2 = -4/5


Now, taking RHS

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

$= \frac{ - 4}{5}$


Hence in both the case LHS = RHS
so, it's Verified !!

Answer 2

$Heya$‼

$Given\:Polynomial:5x^2-4-8x$

As we know,
the standard form of a quadratic polynomial is $ax^2+bx+c$

So, $5x^2-4-8x$
$=>5x^2-8x-4=0$
$=>5x^2+2x-10x-4=0$
$=>x(5x+2)-2(5x+2)=0$
$=>(x-2)(5x+2)=0$
$=>x=2\:or\:x=\frac{-2}{5}$


Let @ and ß be the zeroes of the given polynomial.


Now,

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

OR

$=$@ + ß


$=-\frac{-8}{5}$

OR

$=2-\frac{2}{5}$


$=>\frac{8}{5}$



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

OR

$=$@ * ß


$=\frac{-4}{5}$

OR

$=2*\frac{-2}{5}$


$=>\frac{-4}{5}$