show that the quadratic 2 X square + 5 root 3 X + 6 has real root and always solve it
Answers
Answered by
165
Answer:
Step-by-step explanation:
2x^2 + 5√3x+6 = 0
Here a=2, b = 5√3< c=6
Fir a quadratic equation to have real roots
b^2- 4ac > or = 0
(5√3)^2 - 4 ×2 ×6
= 25 × 3 -48
=75 - 48
=27 > 0
Hence real roots exist
Now finding the roots
Splitting the middle term
2x^2 + 4√3x + √3 x + 6 =0
2x( x +2√3) + √3 (x + 2√3) = 0
(2x+√3)(x+2√3)=0
So roots are
x = -√3/2, x= -2√3
Answered by
22
step by step explanation;
2x^2 + 5√3x+6 = 0
Here a=2, b = 5√3< c=6
Fir a quadratic equation to have real roots
b^2- 4ac > or = 0
(5√3)^2 - 4 ×2 ×6
= 25 × 3 -48
=75 - 48
=27 > 0
Hence real roots exist
Now finding the roots
Splitting the middle term
2x^2 + 4√3x + √3 x + 6 =0
2x( x +2√3) + √3 (x + 2√3) = 0
(2x+√3)(x+2√3)=0
So roots are
x = -√3/2, x= -2√3
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