Find the roots of the quadratic equation given in Q.1 by applying the quadratic formula.?
Answers
Step-by-step explanation:
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Answer:
For a given quadratic equation is ax2 + bx + c = 0,
If b2 - 4ac ≥ 0, then the roots are x = [-b ±√(b2 - 4ac)]/2a
If b2 - 4ac < 0, then no real roots exist.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula. [(i) 2x^2 - 7x + 3 = 0, (ii) 2x^2 + x - 4 = 0, (iii) 4x^2 + 4√3x + 3 = 0, (iv) 2x^2 + x + 4 = 0]
(i) 2x2 - 7x + 3 = 0
a = 2, b = - 7, c = 3
b2 - 4ac = (-7)2 - 4(2)(3)
= 49 - 24 = 25
b2 - 4ac > 0
∴ Roots are x = [- b ± √(b2 - 4ac)]/2a
x = [-(- 7) ± √(25)]/2(2)
x = (7 ± 5)/4
x = (7 + 5)/4 and x = (7 - 5)/4
x = 12/4 and x = 2/4
x = 3 and x = 1/2
Roots are 3, 1/2
(ii) 2x2 + x - 4 = 0
a = 2, b = 1, c = -4
b2 - 4ac = (1)2 - 4(2)(-4)
= 1 + 32 = 33
b2 - 4ac > 0
∴ Roots are x = [- b ± √(b2 - 4ac)]/2a
= (- 1 ± √33)/2(2)
= (- 1 ± √33)/4
x = (-1 + √33)/4 and x = (- 1 - √33)/4
Roots are: (- 1 + √33)/4, (- 1 - √33)/4
(iii) 4x2 + 4√3 x + 3 = 0
a = 4, b = (4√3), c = 3
b2 - 4ac = (4√3)2 - 4(4)(3)
= 48 - 48 = 0
b2 - 4ac = 0
∴ Roots are x = [- b ± √(b2 - 4ac)]/2a
= [- b ± 0]/2a
= - b/2a
= - 4√3 / 2(4)
= - √3/2
Roots are - √3/2, - √3/2
(iv) 2x2 + x + 4 = 0
a = 2, b = 1, c = 4
b2 - 4ac = (1)2 - 4(2)(4)
= 1 - 32 = - 31
b2 - 4ac < 0
∴ No real roots exist.
Step-by-step explanation: