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प्रश्न
Solve the following quadratic equation by formula method
`sqrt(2)"f"^2 - 6"f" + 3sqrt(2)` = 0
योग
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उत्तर
Here a = `sqrt(2)`, b = – 6 and c = `3sqrt(2)`
x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`
= `(6 ± sqrt(36 - 4(sqrt(2))(3sqrt(2))))/(2sqrt(2))`
= `(6 ± sqrt(36 - 24))/(2sqrt(2))`
= `(6 ± sqrt(12))/(2sqrt(2))`
= `(6 ±2sqrt(3))/(2sqrt(2))`
= `(2[3 ±sqrt(3)])/(2sqrt(2))`
= `(3 ± sqrt(3))/sqrt(2)`
The solution set is `(3 + sqrt(3))/sqrt(2)` and `(3 - sqrt(3))/sqrt(2)`
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क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Algebra - Exercise 3.11 [पृष्ठ ११४]
