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The sum of two fractions is 11/12 and their product is 1/8. What is the lesser of the two fractions? Express your answer as a common fraction.

Jun 1, 2021

#1
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a + b  =  11/12    ⇒   b  =  11/12  - a      (1)

ab  =  1/8       (2)          sub (1)  into 2

a (11/12   - a)  =  1/8

-a^2  + (11/12)a  -  1/8  =  0         multiply  through  by  -1

a^2  - (11/12)a  + 1/8   =  0          mulyiply through  by  24

24a^2  - 22a  +  3   =  0             factor as

(6a  -1) ( 4a - 3)   = 0

Setting each  factor to  0  and solving for  a  gives us

a = 1/6        or      a  = 3/4

The smaller fraction  is     1 /  6   Jun 1, 2021
#2
+2

Given a quadratic $ax^2 + bx + c,$

$- \frac{b}{a} = \frac{11}{12}$

$\frac{c}{a} = \frac{1}{8}$

$a = \text{LCM}(12, 8) = 24$

$- \frac{b}{24} = \frac{11}{12}$

$-b = 22$

$b = -22$

$\frac{c}{24} = \frac{1}{8}$

$c = 3$

$24x^2 - 22x + 3 = 0$

$x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4 \cdot 24 \cdot 3}}{24 \cdot 2} = \frac{22 \pm \sqrt{484 - 288}}{48} = \frac{22 \pm \sqrt{196}}{48} = \frac{22 \pm 14}{48} = \frac{11 \pm 7}{24}.$

$x = \frac{18}{24} = \frac{3}{4}$

$x = \frac{4}{24} = \frac{1}{6}$

$\boxed{\frac{1}{6}}$ is the answer edited by MathProblemSolver101  Jun 1, 2021