Question

question no.10 please answer anyone fast​

Answer 1

2x^2-3+5x

2x^2+5x-3 2×3=6

2x^2-6x+1x-3 6=-6×1

2x(1-3) +1(x-3)

(2x+1) (x-3)

2x+1=0

2x=-1

x=-1/2

(x-3)

x-3=0

x=3

x=-1/2 x=3 are the zeros

Answer 2

HEY MATE YOUR ANSWER IS HERE

FINDING ITS ZEROES

${2x}^{2} - 3 + 5x$

OR IN STANDARD FORM

${2x}^{2} + 5x - 3$

BY MIDDLE TERM SPLIT....

${2x}^{2} + ( 6x - x) - 3$

THEN

${2x}^{2} + 6x - x - 3$

HENCE

$2x(x + 3) - 1(x + 3)$

SO

$(2x - 1) \: and \: (x + 3)$

THEREFORE VALUE OF ZEROS IS

$x = \frac{1}{2} \: (be \: \alpha )$

$x = - 3 \: (be \ \: \beta \: )$

now

$sum \: of \: zeroes \: = \frac{ - b}{a}$

$product \: of \: zeroes \: = \frac{c}{a}$

NOW BY VERIFICATION

$\frac{1}{2} - 3 = - (\frac {5}{2} )$

HENCE

$\frac{1 - 6}{2} = \frac{ - 5}{2}$

HENCE

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

NOW PRODUCT..

$( - 3)( \frac{1}{2} ) = \frac{ - 3}{2}$

SO

$\frac{ - 3}{2} = \frac{ - 3}{2}$

HOPE IT HELPS.. THANKS FOR THE QUESTION ☺️☺️☺️☺️

AND REALLY SORRY FOR DELAYING....