State whether the quadratic equation $x^{2}-3x+4=0$ has two distinct real roots. Justify your answer.
(N/A) Given equation is $x^{2}-3x+4=0$.
On comparing this with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-3, c=4$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
Substituting the values:
$D = (-3)^{2} - 4(1)(4)$
$D = 9 - 16$
$D = -7$.
Since $D < 0$,the quadratic equation $x^{2}-3x+4=0$ has no real roots. Therefore,it does not have two distinct real roots.
On comparing this with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-3, c=4$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
Substituting the values:
$D = (-3)^{2} - 4(1)(4)$
$D = 9 - 16$
$D = -7$.
Since $D < 0$,the quadratic equation $x^{2}-3x+4=0$ has no real roots. Therefore,it does not have two distinct real roots.
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