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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 weeks ago

Solve the system: y=14x2+2x−6 y=−14x2−3x+6?

The parabolas above intersect in two places, at (a , b) and (c , d), where a, b, c, and d are all integers. a + b + c + d =

7 Answers

Relevance
  • 2 weeks ago

    x = -3/4, y = 3/8

    x = 4/7, y = -2/7

  • 2 weeks ago

    Subtract the two eq'ns to eliminate 'y' 

    Hence 

    0 = 28x^2 + 5x -12 

    Apply the Quadratic eq'n 

    x = {- 5 +/- sqrt[(5)^2 - 4(28)(-12)]} / 2(28)

    x = { - 5 +/- sqrt[25 + 1344]} / 56

    x = { - 5 +/- sqrt[1369]} / 56 

    x = { - 5 +/- 37}/56 

    x = -42/56 = - 3/4(a) & 32/56 = 4/7(c)

    Substitute for 'y' 

    y = 14((-3/4)^2 + 2(-3/4) - 6 

    y = 7 7/8 - 3/2 - 6 

    y = 3/8 (b) 

    &

    y = 14(4/7)^2 - 2(4/7) - 6 

    y = 4 4/7 - 8/7 - 6 

    y = -2 4/7 (d) 

    Hence a +b+c+d = -3/4 +3/8 +4/7  -2 4/7 = -2 3/8 

  • 2 weeks ago

    y=14x^2+2x-6-------(1)

    y=-14x^2-3x+6-------(2)

    (1)+(2)

    =>

    y=-x/2------(3)

    combining (1) & (3), get

    -x/2=14x^2+2x-6

    =>

    28x^2+5x-12=0

    =>

    x=-0.75~-1 (a)

    y=0.375~0 correspondingly (b)

    or

    x=0.5714286~1 (c)

    y=-0.2857143~0 (d)

    =>

    a+b+c+d=-1+0+1+0=0

    approximately.

  • 2 weeks ago

    y = 14x² + 2x - 6

    y = - 14x² - 3x + 6

    y = y

    14x² + 2x - 6 = - 14x² - 3x + 6

    28x² + 5x = 12

    x² + (5/28).x = 12/28

    x² + (5/28).x + (5/56)² = (12/28) + (5/56)²

    [x + (5/56)]² = 37²/56²

    x + (5/56) = ± 37/56

    x = - (5/56) ± (37/56)

    x = (- 5 ± 37)/56

    First case: x = (- 5 + 37)/56 = 32/56 = 4/7

    y = 14x² + 2x - 6

    y = 14.(4/7)² + 2.(4/7) - 6

    y = (32/7) + (8/7) - (42/7)

    y = - 2/7

    → First point (4/7 ; - 2/7)

    Second case: x = (- 5 - 37)/56 = - 42/56 = - 3/4

    y = 14x² + 2x - 6

    y = 14.(- 3/4)² + 2.(- 3/4) - 6

    y = (63/8) - (12/8) - (48/8)

    y = 3/8

    → Second point (- 3/4 ; 3/8)

  • Alan
    Lv 7
    2 weeks ago

    14x^2  + 2x  -6  = -14x^2 -3x + 6 

    28x^2  + 5x -12 =  0  

    use quadratic formula

    x=  ( -5  +/- sqrt( 25 - 4*(-12)*(28) ) ) / 56  

    x = (-5 +/- sqrt( 1369)  ) / 56  

    x = (-5 +/- 37 ) /56  

    x  =  (-42/56) or   (32/56)   

    simplifying  

    42/14  =3 ,  56/14 = 4

    32/8 =4 ,    56/8  = 7

    x =  -3/4 or    4/7      

    y= 14*(-3/4)^2 +  2*(-3/4) - 6 = 14*(9/16) -6/4  -6  

    y = 7*9/8 -6/4  -6   =  (63/8) - 12/8  - 48/8 =    3/8  

    so one point is 

    (-3/4,   3/8)  

    2nd point  

    y = 14*(4/7)^2   + 2*(4/7)  -6  =   14*(16/49)  +8/7-  42/7 

    y =  (2*16)/7 +8/7 -42/7  =   40/7  -42/7 = -2/7   

    2nd point 

    (4/7,   -2/7)

    a+b + c + d = (-3/4) + (3/8)  + 4/7-2/7 = 

    a+ b + c + d =  -6/8 + 3/8    + 2/7  =   -3/8  + 2/7 

    a+ b + c + d  =    -21/56   +    16/56   =   -5/56 

    a + b + c + d =  -5/56  

    approx.  - -0.089285714

  • 2 weeks ago

    y = 14x^2 + 2x − 6 

    y = −14x2 − 3x + 6

    14x^2 + 2x − 6 = −14x2 − 3x + 6

    28x^2 + 5x - 12 = 0

    (4 x + 3) (7 x - 4) = 0

    Solutions: 

    x = -3/4, y = 3/8

    x = 4/7, y = -2/7

    a + b + c + d = -3/4 + 3/8 + 4/7 - 2/7 = -5/56

  • 2 weeks ago

    Presuming these are:

    y = 14x² + 2x - 6 and y = -14x² - 3x + 6

    You have a system of two equations with two unknowns.  We can solve this with substistution.

    Both expressions are equal to y so both expressions are equal to each other:

    14x² + 2x - 6 = -14x² - 3x + 6

    Now we can solve for x:

    28x² + 5x - 12 = 0

    I don't like trying to factor quadratics with such a high leading coefficient, so I'll use the quadratic equation:

    x = [ -b ± √(b² - 4ac)] / (2a)

    x = [ -5 ± √(5² - 4(28)(-12))] / (2 * 28)

    x = [ -5 ± √(25 + 1344)] / 56

    x = (-5 ± √1369) / 56

    x = (-5 ± 37) / 56

    x = -42/56 and 32/56

    x = -3/4 and 4/7

    These aren't integers, but I'll finish this, anyway.

    We have two values of x so we can find two values of y:

    y = 14x² + 2x - 6

    y = 14(-3/4)² + 2(-3/4) - 6 and y = 14(4/7)² + 2(4/7) - 6

    y = 14(9/16) - 3/2 - 6 and y = 14(16/49) + 8/7 - 6

    y = 126/16 - 3/2 - 6 and y = 224/49 + 8/7 - 6

    y = 63/8 - 3/2 - 6 and y = 224/49 + 8/7 - 6

    y = 63/8 - 12/8 - 48/8 and y = 224/49 + 56/49 - 294/49

    y = 3/8 and y = -14/49

    y = 3/8 and y = -2/7

    The solutions are:

    (-3/4, 3/8) and (4/7, -2/7)

    The sum of the four fractions is:

    -3/4 + 3/8 + 4/7 - 2/7

    -6/8 + 3/8 + 4/7 - 2/7

    -3/8 + 2/7

    -21/56 + 16/56

    -5/56

    Attachment image
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