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What are the roots of this quartaic equation? -10x^2+12x-9=0

User Nabuchodonozor
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1 Answer

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22 votes

Answer:

x=(3/5) + (3√6)i/10, and x =(3/5) - (3√6)i/10

Explanation:

Use the quadratic formula for ax² +bx +c=0

x= [-b±√(b²-4ac)]/2a

For, -10x² +12x -9 = 0

x= [-12 ±√(12²-4(-10)(-9)] /2(-10)

x= [-12 ±√(144-360)] /-20

x= [-12 ±√(144-360)] /-20

x= (-12 ±i√216) /-20

x= (-12 ±i6√6) /-20

so one solution will be

x= (-12 +i6√6) /-20

= (3/5) - (3√6)i/10

the second solution will be

x= (-12 - i6√6) /-20

= (3/5) + (3√6)i/10

User Mohammad Adnan
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