Question

How many and what type of solutions does 7x^2−4x+3 have? 1 rational solution 2 rational solutions 2 irrational solutions 2 nonreal solutions

Answer 1

Solution given:

7x²-4x+3=0

taking common 7

7(x²+4/7*x+3/7)=0

x²+2*x*2/7+(2/7)²-(2/7)²+3/7=0

(x+2/7)²=-17/49

x+2/7=
`\sqrt{(17)/(49)}`

x=±√(17)/7-2/7

taking positive

x=√(17)/7-2/7

x=0.39

taking positive

x=-√(17)/7-2/7

x=-0.87

2:unreal number

Answer 2

2 non real solutions.

Explanation:

We need to use discriminant,

for ax²+bx+c=0

The discriminat is b²-4ac

If the discriminant is,

→ less than 0, then 0 real solutions

→ equal to 0, then 1 real solutions

→ more than 0, then 2 real solutions

Given that,

7x²−4x+3=0

a=7, b=-4, and c=3

→ (-4)²-4(7)(3)

→ 16-84

→ -68

You can see this is less than 0, then non real solutions. [2 nonreal solutions]