Question

Q)
Find the zeros of the quadratic polynomial x2+7/2x+3/4 , and verify relationship between the zeros and the coefficients.

Answer 1

x² + 7/2x + 3/4 = 0
⇒x² + 7/2x = -3/4
⇒x² + 7/2x + (7/4)² = -3/4 + (7/4)²
⇒ x² + 2.(7/4).x + (7/4)² = -3/4 + 49/16
⇒(x + 7/4)² = (-12 + 49)/16
⇒(x + 7/4)² = 37/16
Taking square root both sides,
⇒x + 7/4 = ± √37/4
⇒ x = -7/4 ± √37/4

Hence, roots are (-7 + √37)/4 and (-7 - √37)/4

Now , some of roots = -coefficient of x/Coefficient of x²
= -(7/2)/1 = -7/2
and { (-7 + √37) + (-7 - √37)}/4 = -7/2

Also product of roots = constant/Coefficient of x² = 3/4
And {-7+√37}{-7-√37}/16 = {7² - √37²}/16 = 12/16 = 3/4

You can see both sum of roots and products are verified above explanation.

Answer 2

Step-by-step explanation:

=X square + 7 by 2 X equal to minus 3 by 4

=8 square + 7 by 2 X + bracket 7 by 4 bracket close to equal to minus 3 54 + bracket 7 by 4 bracket close whole twice

=bracket x + 7 by 4 bracket close whole twice is equal to bracket - 12 + 49 bracket close by 16

=bracket X + 7 by 4 bracket close squa=37 by 16

taking a square by both side

=X + 7 by 4 is equal to plus minus root 37 by14

hence, same or roots=coefficient of x by coefficient x square

=minus bracket 7 by 2 bracket close =minus 7 by 2

={(- 7 + root 37)+(-7 - route 37)}/4=-7/2

also product of roots constant by coefficient of x square is equal to 3 by 4

I hope it is helpful