4x^2-8x+3=0 — step-by-step solution

1) Let's solve in

ℝ

4 x^{2} - 8 x + 3 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 4

b = -8

and

c = 3

Δ = {(-8)}^{2} - 4 \cdot 4 \cdot 3

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-8)}^{2}

(-8) \cdot (-8)

64

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

Δ = 64 - 16 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

16 \cdot 3

(10 + 6) \cdot 3

10 \cdot 3 + 6 \cdot 3

30 + 18

48

Δ = 64 - 48

ⓘ Start from

64

and move back

48

on the number line:

64 - 48

16

Δ = 16

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-8) - \sqrt{16}}{2 \cdot 4}

x_{1} = \frac{8 - \sqrt{16}}{2 \cdot 4}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

ⅰ

2 \cdot 4

8

x_{1} = \frac{8 - 4}{8}

ⓘ The numerator and the denominator are computed before the division:

8 - 4

4

x_{1} = \frac{4}{8}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

4

, their greatest common divisor:

\frac{4}{8}

\frac{1 \cdot 4}{2 \cdot 4}

\frac{1}{2}

x_{1} = \frac{1}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-8) + \sqrt{16}}{2 \cdot 4}

x_{2} = \frac{8 + \sqrt{16}}{2 \cdot 4}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

ⅰ

2 \cdot 4

8

x_{2} = \frac{8 + 4}{8}

ⓘ The numerator and the denominator are computed before the division:

8 + 4

12

x_{2} = \frac{12}{8}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

4

, their greatest common divisor:

\frac{12}{8}

\frac{3 \cdot 4}{2 \cdot 4}

\frac{3}{2}

x_{2} = \frac{3}{2}

▶Solutions found:

x = {\frac{1}{2}, \frac{3}{2}}

2) The numeric value of:

x = {\frac{1}{2}, \frac{3}{2}}

equals:

▷

x = {0.5, 1.5}