-3x(x-7)+57x<-8(-7x+8)-3x

Question

Addition and multiplication of inequalities

 

Answer 1

-3x(x-7)+57x < -8(-7x+8)-3x

-3x^2+21x +57x< 56x-64-3x

Bring all terms to one side.

-3x^2+21x+57x-56x+64+3x < 0

-3x^2+25x+64 <0

3x^2-25x-64 < 0

Compare it to ax^2+bx+c < 0

a = 3, b = -25, c = -64

x < [-b±√(b^2-4ac)]/2a

x < [-(-25)±√[625-4*3*-64]/2*3

x < 25±√(625+768)]/6

x < (25±√1393)/6

Answer 2

The inequality is -3x(x - 7) + 57x < -8(-7x + 8) - 3x.

Write the above inequality in general form (with the polynomial on one side and zero on the other).

-3x² + 21x + 57x < 56x - 64 - 3x

-3x² + 78x < 53x - 64

-3x² + 25x + 64 < 0

3x² - 25x - 64 > 0

To find the key numbers, solve the equation 3x² - 25x - 64 for x.

a = 3, b = - 25 and c = -64.

Quadratic formula : x = [-b ± √(b² - 4ac)]/2a

x = [-(-25) ± √{(-25)² - 4(3)(-64)}]/2(3)

x = [25 ± √1393]/6

x = [25 ± 37.32]/6

x = [25 + 37.32]/6 and x = [25 - 37.32]/6

x = 10.387 and x = -2.054.

Test Interval          x-Value                 Polynomial Value                    Conclusion

(-∞, -2.054)           x = -3              3(-3)² - 25(-3) - 64 = 38                Positive

(-2.054, 10.387)    x = 0                3(0)² - 25(0) - 64 = - 64                Negative

(10.387, ∞)          x = 11               3(11)² - 25(11) - 64 = 24              Positive

From this you can conclude that the inequality is satisfied for all x values in (-∞, -2.054) and (10.387, ∞).

The solution set in interval notation form is (-∞, -2.054) U (10.387, ∞).