Math, asked by sivasai8929, 1 year ago

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

Answers

Answered by ALTAF11
366
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 !!
Answered by Anonymous
127
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}
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