Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
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
- lenpol7Lv 71 week 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
- ?Lv 71 week 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.
- la consoleLv 72 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)
- ?Lv 72 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
- ?Lv 72 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
- llafferLv 72 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