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Maharashtra State BoardSSC (English Medium) 10th Standard

Solve the following quadratic equation. 3x2+2x-23 = 0 - Algebra

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Question

Solve the following quadratic equation.

`sqrt(3) x^2 + sqrt(2)x - 2sqrt(3)` = 0

Sum
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Solution

`sqrt(3) x^2 + sqrt(2)x - 2sqrt(3)` = 0

Comparing the above equation with ax2 + bx + c = 0, we get

a = `sqrt(3)`, b = `sqrt(2)`, c = `-2sqrt(3)`

∴ b2 – 4ac = `(sqrt(2))^2 - 4 xx sqrt(3) xx (-2sqrt(3))`

= 2 + 24

= 26

x = `(-"b" +- sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-sqrt(2)  +- sqrt(26))/(2sqrt(3))`

= `(sqrt(2)(-1 +- sqrt(13)))/(2sqrt(3))`

= `(sqrt(2)(-1 +- sqrt(13)))/(2sqrt(3)) xx sqrt(2)/(sqrt(2)`

= `(2(-1 +- sqrt(13)))/(2sqrt(6))`

∴ x = `(-1 +- sqrt(13))/sqrt(6)`

∴ x = `(-1 + sqrt(13))/sqrt(6)` or x = `(-1 - sqrt(13))/sqrt(6)`

∴ The roots of the given equation are `(-1 + sqrt(13))/sqrt(6)` and x = `(-1 - sqrt(13))/sqrt(6)`.

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Chapter 2: Quadratic Equations - Q.3 (B)

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