Question

the roots of quadratic equation 2x^{2} +3x+1=0
are

Answer 1

Answer:

2x^2+3x+1=0

By splitting the middle term,

a=2,b=3,c=1

product=a*c=2*1=2

sum=b=3

(2,1)

2x^2+2x+1x+1=0

2x(X+1)+1(X+1)=0

(x+1) (2x+1)=0

x+1=0 2x+1=0

x= -1 x= -1/2

Step-by-step explanation:

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Answer 2

Answer:

QUADRATIC FORMULA:

$\frac{ - b \frac{ + }{ - } \sqrt{ {b}^{2} - 4ac} }{2a}$

HERE,

a = 2

b = 3

c = 1

$\frac{ - 3 \frac{ + }{ - } \sqrt{ {3}^{2} - 4 \times 2 \times 1} }{2 \times 2}$

$= \frac{ - 3 \frac{ + }{ - } \sqrt{1} }{4}$

HERE IT CAN GO 2 WAYS. AS ROOT OF 1 CAN BE 1 OR -1.

WAY 1:

$= \frac{ - 3 + 1}{4} = \frac{ - 2}{4} = - 0.5$

WAY 2:

$= \frac{ - 3 - 1}{4} = \frac{ - 4}{4} = - 1$

SO ROOTS OF EQUATION ARE -1 AND -0.5

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