Question

Find the solutions of the quadratic equation 14x^2+9x+10=014x

2
+9x+10=014, x, start superscript, 2, end superscript, plus, 9, x, plus, 10, equals, 0.
Choose 1 answer:
Choose 1 answer:

(Choice A)
A
\dfrac{9}{28}\pm\dfrac{\sqrt{479}}{28}i
28
9
​ ±
28
479
​
​ istart fraction, 9, divided by, 28, end fraction, plus minus, start fraction, square root of, 479, end square root, divided by, 28, end fraction, i

(Choice B)
B
-\dfrac{9}{28}\pm\dfrac{\sqrt{479}}{28}i−
28
9
​ ±
28
479
​
​ iminus, start fraction, 9, divided by, 28, end fraction, plus minus, start fraction, square root of, 479, end square root, divided by, 28, end fraction, i

(Choice C)
C
-\dfrac{9}{28}\pm\dfrac{\sqrt{479}}{28}−
28
9
​ ±
28
479
​
​ minus, start fraction, 9, divided by, 28, end fraction, plus minus, start fraction, square root of, 479, end square root, divided by, 28, end fraction

(Choice D)
D
\dfrac{9}{28}\pm\dfrac{\sqrt{479}}{28}
28
9
​ ±
28
479
​
​

Answer 1

Option B.
`x=-(9)/(28)(+/-)(√(479))/(28)i`

Explanation:

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

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

in this problem we have

`14x^(2)+9x+10=0`

so

`a=14\\b=9\\c=10`

substitute in the formula

`x=\frac{-9(+/-)\sqrt{9^(2)-4(14)(10)}} {2(14)}`

`x=\frac{-9(+/-)√(-479)} {28}`

Remember that

`i=√(-1)`

substitute

`x=\frac{-9(+/-)√(479)i} {28}`

`x=-(9)/(28)(+/-)(√(479))/(28)i`