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Maharashtra State BoardSSC (English Medium) 10th Standard

Compare the quadratic equation x2+93x+24=0 to ax2 + bx + c = 0 and find the value of discriminant and hence write the nature of the roots.

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Question

Compare the quadratic equation `x^2 + 9sqrt(3)x + 24 = 0` to ax2 + bx + c = 0 and find the value of discriminant and hence write the nature of the roots.

Sum
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Solution

Given equation is, `x^2 + 9sqrt(3)x + 24 = 0`

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

a = 1, b = `9sqrt(3)`, c = 24

Discriminant, D = b2 – 4ac

= `(9sqrt(3))^2 - 4(1)(24)`

= 243 – 96

= 147

Here, D > 0.

As a result, the roots of the quadratic equation are both real and unequal.

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Solution :

Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0

a = 1, b = 2, c = `square`

∴ b2 – 4ac = (2)2 – 4 × `square` × `square`

Δ = 4 + `square` = 40

∴ b2 – 4ac > 0

∴ The roots of the equation are real and unequal.


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