Explanation
Step 1: Find the coefficients $$\boldsymbol{a,\ b, \text{and}\ c}$$
Compare the given equation with $$ax^2+bx+c=0$$, we get
$$a=3,\ b=-2\sqrt{6},\ c=2$$
Step 2: Find discriminant of the given quadratic equation
$$D=b^2-4ac=(-2\sqrt{6})^2-4(3)(2)=12-12=0$$
Hence the roots of given quadratic equation are real and equal.
Hence the given statement is true.
Let $$z$$ be a complex number such that $$\left| z+\dfrac { 1 }{ z } \right| =2$$.
If $$\left| z \right| ={ r }_{ 1 }$$ and $$\left| \dfrac { 1 }{ z } \right| =$$ $${r}_{2}$$ for $$\arg z=\dfrac { \pi }{ 4 }$$ then
$$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =$$
If the value of '$$b^2-4ac$$'is equal to zero, the quadratic equation $$ax^2+bx+c=0$$ will have
A particle starts from a point $$z_0= I + i$$, where $$i=\sqrt{-1}$$ It moves horizontally away from origin by $$2$$ units and thenvertically away from origin by $$3$$ units to reach a point$$ z_1$$. From $$z_1$$particle moves $$\sqrt{5}$$ units in the direction of $$2\hat i + \hat j$$ andthen it moves through an angle of $$\cos e{c^{ - 1}}\sqrt 2 $$ in anticlockwisedirection of a circle with centre at origin to reach a point $$z_2$$ . The arg $$z_2$$ is given by
(A) Which of the following quadratic polynomials has zeros-9 and 9
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