Question

Find the following values of the discriminant for the following quadratic equations ​d =b^2-4ac methods camplete solution

Answer 1

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Answer 2

SolutioN :

i ) x² + 10x - 7 = 0.

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c = 0.$

Where as,

• a = 1.
• b = 10.
• c = - 7.

D = b² - 4ac.

→ D = ( 10 )² - 4 ( 1 ) ( - 7 )

→ D = 100 - 28.

→ D = 128.

More InformatioN :

We have, Quadratic Formula.

$\tt \dagger \: \: \: \: \: \boxed{ \tt x = \dfrac{ - b\pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\tt : \implies \tt x = \dfrac{ - 10\pm \sqrt{ {(10)}^{2} - 4 \times 1 \times - 7 } }{2}$

$\tt : \implies \tt x = \dfrac{ - 10\pm \sqrt{ 100 + 28 } }{2}$

$\tt : \implies \tt x = \dfrac{ - 10\pm \sqrt{ 1 28 } }{2}$

$\tt : \implies \tt x = \dfrac{ - 10 + \sqrt{ 1 28 } }{2}$

$\tt : \implies \tt x = \dfrac{ - 10 - \sqrt{ 1 28 } }{2}$

$\rule{90}2$

( ii ) 2x² - 3x + 5 = 0.

Where as,

• a = 2.
• b = - 3.
• c = 5.

→ D = b² - 4ac.

→ D = ( - 3 )² - 4( 2 )(5).

→ D = 9 - 10 * 4.

→ D = 9 - 40.

→ D = - 31.

MorE InformatioN :

$\tt : \implies \tt x = \dfrac{ 3\pm \sqrt{ {( - 3)}^{2} - 4 \times 2 \times 5 } }{2 \times 2}$

$\tt : \implies \tt x = \dfrac{ 3\pm \sqrt{ 9 - 40 } }{2 \times 2}$

$\tt : \implies \tt x = \dfrac{ 3\pm \sqrt{ - 31 } }{4}$

( iii ) 2m² + m - 1 = 0.

Where as,

• a = 2.
• b = 1.
• c = - 1.

→ D = b² - 4ac.

→ D = ( 1 )² - 4(2)(-1)

→ D = 1 + 8.

→ D = 9.

$\rule{90}2$

( iv ) 3x² - 4√3x + 4 = 0.

Where as,

• a = 3.
• b = - 4√3.
• c = 4.

→ D = b² - 4ac.

→ D = ( - 4√3 )² - 4(3)(4).

→ D = 48 - 48.

→ D = 0.

$\rule{90}2$

( v ) x² + 7x + 6 = 0.

Where as,

• a = 1.
• b = 7.
• c = 6.

D = b² - 4ac.

→ D = ( 7 )² - 4(1)(6).

→ D = 49 - 24.

→ D = 25.

$\rule{90}2$