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

Question

contestada

Respuesta :

calculista calculista · Answer 1 · 2019-06-01T03:48:48+02:00

Answer:

Option B. [tex]x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i[/tex]

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to

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

in this problem we have

[tex]14x^{2}+9x+10=0[/tex]

so

[tex]a=14\\b=9\\c=10[/tex]

substitute in the formula

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

[tex]x=\frac{-9(+/-)\sqrt{-479}} {28}[/tex]

Remember that

[tex]i=\sqrt{-1}[/tex]

substitute

[tex]x=\frac{-9(+/-)\sqrt{479}i} {28}[/tex]

[tex]x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i[/tex]