Question

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

Answer 1

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

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

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

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

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

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

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

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

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

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

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Answer 2

Step-by-step 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