Question

1) Determine the discriminant of the 2nd degree equation below:

3x 2 − 2x − 1 = 0
a = 3, b = −2, c = −1
Discriminant → ∆= b 2 − 4 a c


2) Solve the following 2nd degree equations using Bháskara's formula:

Δ = b² - 4.a.c
x = - b ± √Δ
__________
2a

a) x 2 + 5x + 6 = 0

b)x 2 + 2x + 1 = 0

c) x2 - x - 20 = 0

d) x2 - 3x -4 = 0

Answer 1

Explanation:

a)

given: a = 1, b = 5, c = 6

1) Discriminant → ∆= b² − (4*a*c)

∆= b² - (4*a*c)

∆= 5² - (4*1*6)

∆=25 - ( 24 )

∆= 25 - 24

∆= 1

2)

Solve x = (- b ± √Δ ) / 2a

x = ( 5 ± √25 ) / 2*1

x = ( 2 ± 5 ) / 2

x = ( 2 + 5 ) / 2 or x = ( 2 - 5 ) / 2

x = ( 7 ) / 2 or x = ( - 3 ) / 2

x = 3.5 or x = -1.5

b)

given: a = 1, b = 2, c = 1

1) Discriminant → ∆= b² − (4*a*c)

∆= b² - (4*a*c)

∆= 2² - (4*1*1)

∆= 4 - (4)

∆= 4 - 4

∆= 0

2)

Solve x = (- b ± √Δ ) / 2a

x = ( -2 ± √0) / 2*1

x = ( 2 ± 0 ) / 2

x = ( 2 + 0) / 2 or x = ( 2 - 0 ) / 2

x = ( 2 ) / 2 or x = ( 2 ) / 2

x = 1 or x = 1

x = 1 (only one solution)

c)

given: a = 1, b = -1, c = -20

1) Discriminant → ∆= b² − (4*a*c)

∆= b² - (4*a*c)

∆= -1² - (4*1*-20)

∆= 1 - ( -80 )

∆= 1 + 80

∆= 81

2)

Solve x = (- b ± √Δ ) / 2a

x = ( 2 ± √81 ) / 2*1

x = ( 2 ± 9 ) / 2

x = ( 2 + 9 ) / 2 or x = ( 2 - 9 ) / 2

x = ( 11 ) / 2 or x = ( - 7 ) / 2

x = 5.5 or x = -3.5

d)

given: a = 1, b = -3, c = -4

1) Discriminant → ∆= b² − (4*a*c)

∆= b² - (4*a*c)

∆= -3² - (4*1*-4)

∆= 9 - ( -16)

∆= 9 + 16

∆= 25

2)

Solve x = (- b ± √Δ ) / 2a

x = ( 3 ± √25 ) / 2*1

x = ( 3 ± 5 ) / 2

x = ( 3 + 5 ) / 2 or x = ( 3 - 5 ) / 2

x = ( 8 ) / 2 or x = ( - 2 ) / 2

x = 4 or x = -1

Answer 2

`\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}`

We have, Discriminant formula for finding roots:

`\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^(2) - 4ac} }{2a} }}}`

Here,

• x is the root of the equation.
• a is the coefficient of x^2
• b is the coefficient of x
• c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

`\large{ \rm{ \longrightarrow \: x = \frac{ - 5\pm \sqrt{( - 5) {}^(2) - 4 * 1 * 6 }} {2 * 1}}}`

`\large{ \rm{ \longrightarrow \: x = ( - 5 \pm √(25 - 24) )/(2 * 1) }}`

`\large{ \rm{ \longrightarrow \: x = ( - 5 \pm 1)/(2) }}`

So here,

`\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}`

❒ p(x) = x^2 + 2x + 1 = 0

`\large{ \rm{ \longrightarrow \: x = \frac{ - 2 \pm \sqrt{ {2}^(2) - 4 * 1 * 1} }{2 * 1} }}`

`\large{ \rm{ \longrightarrow \: x = ( - 2 \pm √(4 - 4) )/(2) }}`

`\large{ \rm{ \longrightarrow \: x = ( - 2 \pm 0)/(2) }}`

So here,

`\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}`

❒ p(x) = x^2 - x - 20 = 0

`\large{ \rm{ \longrightarrow \: x = \frac{ - ( - 1) \pm \sqrt{( - 1) {}^(2) - 4 * 1 * ( - 20) } }{2 * 1} }}`

`\large{ \rm{ \longrightarrow \: x = ( 1 \pm √(1 + 80) )/(2) }}`

`\large{ \rm{ \longrightarrow \: x = (1 \pm 9)/(2) }}`

So here,

`\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}`

❒ p(x) = x^2 - 3x - 4 = 0

`\large{ \rm{ \longrightarrow \: x = \frac{ - ( - 3) \pm \sqrt{( - 3) {}^(2) - 4 * 1 * ( - 4) } }{2 * 1} }}`

`\large{ \rm{ \longrightarrow \: x = (3 \pm √(9 + 16) )/(2 * 1) }}`

`\large{ \rm{ \longrightarrow \: x = (3 \pm 5)/(2) }}`

So here,

`\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}`

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