Question

Find the roots


`\sf x^2+3x+5=0`

Use quadratic formula

Note:-

Plagarised/spam/short answers will be deleted on the spot.

Answer with all steps and proper explanation .

Hint:-

irrational roots will come ​

Answer 1

`⇒x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^(2) } - 4ac }{2a}`

Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

`⇒ {x}^(2) + 3x + 5 = 0 \\ ⇒a = 1 \\ ⇒b = 3 \\ ⇒c = 5 \\ ⇒x = \frac{ - 3 \binom{ + }{ - } \sqrt{ {3}^(2) } - 4.1.5}{2.1}`

▪Evaluate the exponent

▪Multiply the numbers

▪Subtract the numbers

▪Multiply the numbers

`⇒x = \frac{ - 3 \binom{ + }{ - } √( - 11) }{2}`

▪The square root of a negative number is not a real number

`⇒d = - 11`

♧ no solution ♧

Answer 2

`\sf \: {x}^(2) + 3x + 5 = 0`

Solve the quadratic equation (ax²+bx+c=0) using the quadratic formula:

`\boxed{ \bf \: x = \frac{ - b± \sqrt{ {b}^(2) - 4ac } }{2a} }`

• a = 1
• b = 3
• c = 5

`\tt \: x = \frac{ - 3± \sqrt{ {3}^(2) - 4 * 1 * 5 } }{2 * 1}`

Any expression multiplied by 1 remains the same

`\tt \: x = \frac{ - 3± \sqrt{ {3}^(2) - 4 * 5 } }{2}`

Evaluate the power

`\tt \: x = ( - 3± √(9 - 4 * 5) )/(2)`

Multiply the numbers

`\tt \: x = ( - 3± √(9 - 20) )/(2)`

Calculate the difference

`\tt \: x = ( - 3± √( - 11) )/(2)`

Calculate the square root

`\tt \: x = ( - 3± √(11)i )/(2)`

Write the solutions, one with a + sign and one with a - sign

`\tt \: x = ( - 3 + √(11) i)/(2) \\ \tt \: x = ( - 3 - √(11)i )/(2)`

Separate the real and the imaginary parts

`\tt \: x = - (3)/(2) + ( √(11) )/(2) i \\ \tt \: x = - (3)/(2) - ( √(11) )/(2) i`

∴ The quadratic equation has two solutions

`\rm x_(1) = - (3)/(2) + ( √(11) )/(2) i, x_(2) = - (3)/(2) - ( √(11) )/(2) i`