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_1particle 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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