How do you find the discriminant of $2 x^{2} - 3 x + 1 = 0$ $2x^2-3x+1=0$ and use it to determine if the equation has one, two real or two imaginary roots?

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Answer 1 · 2017-08-01T05:48:16

Two real roots $\frac{1}{2}$ $1/2$ and $1$ $1$ .

The discriminant of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is $b^{2} - 4 a c$ $b^2-4ac$ .

Assuming $a$ $a$ , $b$ $b$ and $c$ $c$ are real numbers,

if $b^{2} - 4 a c > 0$ $b^2-4ac>0$ , we have two real roots

if $b^{2} - 4 a c = 0$ $b^2-4ac=0$ , we have one real root, and

if $b^{2} - 4 a c < 0$ $b^2-4ac<0$ , we have two complex conjugate numbers as roots.

For $2 x^{2} - 3 x + 1 = 0$ $2x^2-3x+1=0$ , as $a = 2$ $a=2$ , $b = - 3$ $b=-3$ and $c = 1$ $c=1$ ,

the discriminant is ${(- 3)}^{2} - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 > 0$ $(-3)^2-4*2*1=9-8=1>0$ ,

hence we should have two real roots

Now $2 x^{2} - 3 x + 1 = 0$ $2x^2-3x+1=0$ can be written as

$2 x^{2} - 2 x - x + 1 = 0$ $2x^2-2x-x+1=0$

or $2 x (x - 1) - 1 (x - 1) = 0$ $2x(x-1)-1(x-1)=0$ or $(2 x - 1) (x - 1) = 0$ $(2x-1)(x-1)=0$

i.e. $x = \frac{1}{2}$ $x=1/2$ or $1$ $1$ , i.e. two real roots.

How do you find the discriminant of 2x2−3x+1=02x^2-3x+1=0 and use it to determine if the equation has one, two real or two imaginary roots?