Which of the following has real roots? निम्नलिखित में से किस समीकरण के मूल वास्तविक हैं ? (1 )x2 + x + 1 = 0
(2)x2 + 6x+4= 0
(3) 2x2 +5x+5= 0
(4)x2 + 2x + 3 = 0
Answers
Given : समीकरण
(1 )x² + x + 1 = 0
(2)x² + 6x+4= 0
(3) 2x² +5x+5= 0
(4)x²+ 2x + 3 = 0
Solution:
ax² + bx + c = 0
समीकरण के मूल वास्तविक हैं
real roots if D ≥ 0
D = b² - 4ac
(1 ) x² + x + 1 = 0
a = 1 , b = 1 , c = 1
D = 1² - 4(1)(1) = - 3 < 0
Roots are not real
(2) x² + 6x+4= 0
a = 1 , b = 6 , c = 4
D = 6² - 4(1)(4) = 20 > 0
Roots are real
(3) 2x² +5x+5= 0
a = 2 , b =- 5 , c = 5
D = (-5)² - 4(2)(5) = - 15 < 0
Roots are not real
(4) x²+ 2x + 3 = 0
a = 1 , b = 2 , c = 3
D = 2² - 4(1)(3) = - 8 < 0
Roots are not real
Hence x² + 6x+4= 0 has real roots
x² + 6x+4= 0 समीकरण के मूल वास्तविक हैं
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Solution :-
Concept used :- If A•x^2 + B•x + C = 0 ,is any quadratic equation, then its discriminant is given by;
- D = B^2 - 4•A•C
- If D = 0 , then the given quadratic equation has real and equal roots.
- If D > 0 , then the given quadratic equation has real and distinct roots.
- If D < 0 , then the given quadratic equation has unreal (imaginary) roots.
checking all given options ,
(1 )x² + x + 1 = 0
→ D = (1)² - 4 * 1 * 1
→ D = 1 - 4
→ D = (-3)
since, (-3) < 0 .
the given given quadratic equation has unreal (imaginary) roots.
(2) x² + 6x+4= 0
→ D = (6)² - 4 * 1 * 4
→ D = 24 - 16
→ D = 8
since, 8 > 0
the given given quadratic equation has real roots.
(3) 2x² +5x+5= 0
→ D = (5)² - 4 * 2 * 5
→ D = 25 - 40
→ D = (-15)
since, (-15) < 0 .
the given given quadratic equation has unreal (imaginary) roots.
(4) x² + 2x + 3 = 0
→ D = (2)² - 4 * 1 * 3
→ D = 4 - 12
→ D = (-8)
since, (-8) < 0 .
the given given quadratic equation has unreal (imaginary) roots.
Hence, option (2) x² + 6x+4 = 0 has real roots .
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