Find the discriminant of the following quadratic equations and discuss the nature of the roots .
1. $6 x^{2} - 13 x + 6 = 0$
2. $\sqrt{6} x^{2} - 5 x + \sqrt{6} = 0$
3. $24 x^{2} - 17 x + 3 = 0$
4. $x^{2} + 2 x + 4 = 0$
5. $x^{2} + x + 1 = 0$
6. $x^{2} - 3 \sqrt{3} x - 30 = 0$

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Solution

$1) 6 x^{2} - 13 x + 6 = 0$
$D = b^{2} - 4 a c = (- 13)^{2} - 4 \times 6 \times 6 = 169 - 144 = 25 > 0$
Hence, the roots are real and unequal.

$2) \sqrt{6} x^{2} - 5 x + \sqrt{6} = 0$
$D = b^{2} - 4 a c = (- 5)^{2} - 4 \times \sqrt{6} \times \sqrt{6} = 25 - 24 = 1 > 0$
Hence, the roots are real and unequal.

$3) 24 x^{2} - 17 x + 3 = 0$
$D = b^{2} - 4 a c = (- 17)^{2} - 4 \times 24 \times 3 = 289 - 288 = 1 > 0$
Hence, the roots are real and unequal.

$4) x^{2} + 2 x + 4 = 0$
$D = b^{2} - 4 a c = 4 - 4 \times 1 \times 4 = 4 - 16 = - 12 < 0$
Hence, it has no real roots.

$5) x^{2} + x + 1 = 0$
$D = b^{2} - 4 a c = 1 - 4 \times 1 \times 1 = 1 - 4 = - 3 < 0$
Hence, it has no real roots.

$6) x^{2} - 3 \sqrt{3} x - 30 = 0$
$D = b^{2} - 4 a c = (- 3 \sqrt{3})^{2} - 4 \times 1 \times (- 30) = 27 + 120 = 147 > 0$
Hence, the roots are real and unequal.

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Similar questions

Write the discriminant of the following quadratic equations:
(i) $2 x^{2} - 5 x + 3 = 0$ (ii) $x^{2} + 2 x + 4 = 0$
(iii) $(x - 1) (2 x - 1) = 0$ (iv) $x^{2} - 2 x + k = 0, k ϵ R$
(v) $\sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0$ (vi) $x^{2} - x + 1 = 0$