Question

Find the zeroes of quadratic polynomial 2x2 + 7/2x+4/3

Answer 1

Step-by-step explanation:

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.

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Answer 2

Answer:

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.

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