Math, asked by ismail750md, 23 hours ago

Find the roots of the quadratic equation given in Q.1 by applying the quadratic formula.?

Answers

Answered by surajdev35
0

Step-by-step explanation:

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Answered by ananyanaskar28
1

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:

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