Question

find the value of the roots for 2x^2-7x+3=0 and if real roots exist,find them​

Answer 1

2x²-7x+3 = 0

Here, a = 2, b = -7 ,c = 3

So, Discriminant (D) = b²-4 ac

$\implies$ (-7)²-4×2×3

$\implies$ 49-24

$\implies$ 25

Here D > 0

hence, this equation has real and distinct roots.

By quadratic formula

x = -b+-√D/2a

x = -(-7) +- (5)/4

Now, taking +ve sign.

we get

$\alpha$ = 7+5/4

$\implies$ $\frac{12}{4}$

$\implies$ $3$....ans....

and, taking -ve sign

we get,

$\beta$ = 7-5/4

$\implies$ $\frac{2}{4}$

$\implies$ $\frac{1}{2}$...ans....

Answer 2

Answer:

x=1/2 , 3

Step-by-step explanation:

As real roots exists

hence

D=b²-4ac

=(-7)² -4 ×2×3

=49-24

=25

hence d>0

roots are real and exist

2x²-7x+3=0

2x²-6x-x+3=0

2x(x-3) -1(x-3)=0

(2x-1) (x-3)=0

2x-1 =0. x-3=0

x=1/2. x=3