Question

One solution of 21x^2 + bx -4 = 0 is -4/3. Find b and the other solution.​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given that x = - $\frac{4}{3}$ is a solution of the equation, then

Substitute this value into the equation and solve for b

21 (- $\frac{4}{3}$ )² + b (- $\frac{4}{3}$ ) - 4 = 0

21 × $\frac{16}{9}$ - $\frac{4}{3}$ b - 4 = 0

$\frac{112}{3}$ - $\frac{4}{3}$ b - 4 = 0

Multiply through by 3

112 - 4b - 12 = 0

100 - 4b = 0 ( subtract 100 from both sides )

- 4b = - 100 ( divide both sides by - 4 )

b = 25 ← value of b

The equation can now be written as

21x² + 25x - 4 = 0 ← in standard form

with a = 21, b = 25, c = - 4

Use the quadratic formula to solve for x

x = ( - 25 ± $\sqrt{25^2-(4(21)(-4)}$ ) / 42

  = ( - 25 ± $\sqrt{961}$ ) / 42

  = ( - 25 ± 31 ) / 42

x = $\frac{-25-31}{42}$ = $\frac{-56}{42}$ = - $\frac{4}{3}$

or x = $\frac{-25+31}{42}$ =  $\frac{6}{42}$ = $\frac{1}{7}$

The other solution is x = $\frac{1}{7}$