Question

Determine the nature of the roots for the quadratic equation
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Answer 1

Answer:

To determine the nature of roots of quadratic equations (in the form ax^2 + bx +c=0) , we need to caclulate the discriminant, which is b^2 - 4 a c. When discriminant is greater than zero, the roots are unequal and real. When discriminant is equal to zero, the roots are equal and real

Answer 2

$\Large \underline{\underline{\bf Question}}:$

Determine the nature of the roots for the quadratic equation :

• $\sf \sqrt{3}x^2 + \sqrt{2}x - 2\sqrt{3} = 0$

$\Large \underline{\underline{\bf Answer}}:$

$\underline{\underline{\bf Given}}:$

• $\sf \sqrt{3}x^2 + \sqrt{2}x - 2\sqrt{3} = 0$

$\underline{\underline{\bf To \: Find}}:$

• Nature of the roots = ?

$\underline{\underline{\bf Solution}}:$

$\sf \sqrt{3}x^2 + \sqrt{2}x - 2\sqrt{3} = 0$

By comparing with ax² + bx + c = 0, we have :

• $\sf a= \sqrt{3}$
• $\sf b= \sqrt{2}$
• $\sf c= -2\sqrt{3}$

Now, let's find discriminant (D) :

$\large \: \: \: \: \: \: \: \: \: \: \pink{\underline{\boxed{\bf D = b^2 - 4ac}}}$

$\sf : \implies D = (\sqrt{2})^2 - 4\times\sqrt{3} \times (-2\sqrt{3})$

$\sf : \implies D = 2 + 4\times\sqrt{3} \times 2\sqrt{3}$

$\sf : \implies D = 2 + 8\times (\sqrt{3})^2$

$\sf : \implies D = 2 + 8\times 3$

$\sf : \implies D = 2 + 24$

$\sf : \implies D = 26$

$\large \: \: \: \: \: \: \: \: \: \: \pink{\underline{\boxed{\bf D = 26}}}$

Now, we know that :

• When D = 0, then roots are real and equal.
• When D > 0, then roots are real and unequal.
• When D < 0, then there are no real roots.

$\textsf{As, \pink{D (26) > 0}}$

$\textsf{Hence, nature of roots is \pink{real and unequal}.}$