Question

(i) Determine the nature of the roots for the quadratic equation
√3x + √2r-2/3=0.
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Answer 1

Answer:

real root.

Step-by-step explanation:

The given quadratic equation is

\sqrt{3}x^2+\sqrt{2}x-2\sqrt{3}=03x2+2x−23=0

A quadratic equation is ax^2+bx+c=0ax2+bx+c=0.

If b^2-4ac < 0b2−4ac<0 , then the equation have two complex roots.

If b^2-4ac=0b2−4ac=0 , then the equation have equal real roots.

If b^2-4ac > 0b2−4ac>0 , then the equation have two distinct real roots.

In the given equation,

a=\sqrt{3},b=\sqrt{2},c=-2\sqrt{3}a=3,b=2,c=−23

b^2-4ac=(\sqrt{2})^2-4(\sqrt{3})(-2\sqrt{3})=2+24=26b2−4ac=(2)2−4(3)(−23)=2+24=26

Since b^2-4ac > 0b2−4ac>0 , therefore, the given quadratic equation have two distinct real root.

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