6g^2-49g+8=0 — step-by-step solution

1) Let's solve in

ℝ

6 g^{2} - 49 g + 8 = 0

Given the quadratic form

a g^{2} + b g + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 6

b = -49

and

c = 8

Δ = {(-49)}^{2} - 4 \cdot 6 \cdot 8

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

{(-49)}^{2}

(-49) \cdot (-49)

2401

■Multiplications are done before additions and subtractions:

4 \cdot 6

24

Δ = 2401 - 24 \cdot 8

ⓘ Multiplications are done before additions and subtractions:

24 \cdot 8

(20 + 4) \cdot 8

20 \cdot 8 + 4 \cdot 8

160 + 32

192

Δ = 2401 - 192

ⓘ Start from

2401

and move back

192

on the number line:

2401 - 192

2209

Δ = 2209

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

g_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-49) - \sqrt{2209}}{2 \cdot 6}

g_{1} = \frac{49 - \sqrt{2209}}{2 \cdot 6}

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

\sqrt{2209}

47

;

47^{2}

2209

ⅰ

2 \cdot 6

12

g_{1} = \frac{49 - 47}{12}

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

49 - 47

2

g_{1} = \frac{2}{12}

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

2

, their greatest common divisor:

\frac{2}{12}

\frac{1 \cdot 2}{6 \cdot 2}

\frac{1}{6}

g_{1} = \frac{1}{6}

The quadratic formula gives

g_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-49) + \sqrt{2209}}{2 \cdot 6}

g_{2} = \frac{49 + \sqrt{2209}}{2 \cdot 6}

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

\sqrt{2209}

47

;

47^{2}

2209

ⅰ

2 \cdot 6

12

g_{2} = \frac{49 + 47}{12}

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

49 + 47

96

g_{2} = \frac{96}{12}

We simplify the fraction:

\frac{96}{12}

by the common factor:

12

to get the new expression:

8

g_{2} = 8

▶Solutions found:

g = {\frac{1}{6}, 8}

2) The numeric value of:

g = {\frac{1}{6}, 8}

is:

▷

g = {\approx 0.1667, 8}