Question

The roots of 2x2-6x+3=0 will be (a) real, unequal and rational (b) real,unequal and irrational (c) real and equal (d) imaginary roots

Answer 1

(-6)^2 - 4.2.3
36 -24
=12
so,root be real ,unequal and rational

Answer 2

Answer:

The nature of the root of the quadratic equation $2x^{2} -6x+3=0$ are real, unequal, and rational

Step-by-step explanation:

• A quadratic equation is a type of equation whose degree is two, a quadratic equation can be represented as

                            $ax^{2} +bx+c=0$

• the corresponding root or the value of x that satisfies the quadratic equation is given by the formula

                 $x=\frac{-b+\sqrt{b^{2}-4ac } }{2a}$     or $x=\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

From the question, we have given a quadratic equation of the form

$2x^{2} -6x+3=0$

as compared with the standard equation we get

$a=2\\b=-6\\c=3$

substitute these values to get the roots

$x=\frac{6+\sqrt{36-4*2*3} }{4} =2.35\\x=\frac{6-\sqrt{36-4*2*3} }{4} =0.65$

that is the values of x are real, unequal, and rational