Question

Show that the equation $3x^{2}-4x+5=0$ cannot have any real root.​

Answer 1

⭐《ANSWER》

Actually welcome to the concept of the QUADRATIC EQUATIONS

Basically The polynomial of Degree 2 is termed as a Quadratic Equation

of standard form of ax^2 + bx+ c =0

Now , since the degree of the polynomial is 2 , By default it has PARABOLIC CURVE natured Graph on the coordinate axis ,

Thus it is a MONOTONIC Function ( Strictness in increasing or decreasing )

Since it has degree 2 , it will actually have two roots ,

the condition for the roots to be REAL and NON - REAL ( IMAGINARY)

↗is given by concept of the DISCRIMINANT , denoted by letter "D"

D = b^2 - 4ac

as the result follows as

D > 0 , Roots are real and unequal

D=0 , Roots are real and equal

D < 0 , Roots are imaginary

so here applying the formula for $3x^{2}-4×+5=0$

we get as

↗D= (-4)^2 - 4 ( 3) (5)

D = 16 - 60

↗D = -44

〽since the value of D <0 , then we can say that, there exist no real roots for the given quadratic equation. .

Hence proved

Answer 2

standard form of qurdatic equation is ax²+bx+c=0 but a don't equal to 0.
we know that from nature of root is
b²-4ac <0 than the equation have no real root.
3x²-4x+5=0 in comparison of ax²+bx+c=0
a=3, b= -4, c=5
${b}^{2} - 4ac \\ {( - 4)}^{2} - 4 \times 3 \times 5 \\ 16 - 60 \\ - 44$
so we can say that mention equation have no real root.