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Linear Algebra Examples
d=v(x2-x+(y2-y))⋅22
Step 1
Rewrite the equation as v(x2-x+y2-y)⋅22=d.
v(x2-x+y2-y)⋅22=d
Step 2
Step 2.1
Apply the distributive property.
(vx2+v(-x)+vy2+v(-y))⋅22=d
Step 2.2
Simplify.
Step 2.2.1
Rewrite using the commutative property of multiplication.
(vx2-vx+vy2+v(-y))⋅22=d
Step 2.2.2
Rewrite using the commutative property of multiplication.
(vx2-vx+vy2-vy)⋅22=d
(vx2-vx+vy2-vy)⋅22=d
Step 2.3
Apply the distributive property.
vx2⋅22-vx⋅22+vy2⋅22-vy⋅22=d
Step 2.4
Simplify.
Step 2.4.1
Move 22 to the left of vx2.
22⋅(vx2)-vx⋅22+vy2⋅22-vy⋅22=d
Step 2.4.2
Multiply 22 by -1.
22⋅(vx2)-22vx+vy2⋅22-vy⋅22=d
Step 2.4.3
Move 22 to the left of vy2.
22⋅(vx2)-22vx+22⋅(vy2)-vy⋅22=d
Step 2.4.4
Multiply 22 by -1.
22⋅(vx2)-22vx+22⋅(vy2)-22vy=d
22⋅(vx2)-22vx+22⋅(vy2)-22vy=d
Step 2.5
Remove parentheses.
22vx2-22vx+22vy2-22vy=d
22vx2-22vx+22vy2-22vy=d
Step 3
Subtract d from both sides of the equation.
22vx2-22vx+22vy2-22vy-d=0
Step 4
Use the quadratic formula to find the solutions.
-b±√b2-4(ac)2a
Step 5
Substitute the values a=22v, b=-22v, and c=22vx2-22vx-d into the quadratic formula and solve for y.
22v±√(-22v)2-4⋅(22v⋅(22vx2-22vx-d))2(22v)
Step 6
Step 6.1
Simplify the numerator.
Step 6.1.1
Add parentheses.
y=22v±√(-22v)2-4⋅(22(v⋅(22(vx2)-22(vx)-d)))2⋅(22v)
Step 6.1.2
Let u=22(v⋅(22(vx2)-22(vx)-d)). Substitute u for all occurrences of 22(v⋅(22(vx2)-22(vx)-d)).
Step 6.1.2.1
Apply the product rule to -22v.
y=22v±√(-22)2v2-4⋅u2⋅(22v)
Step 6.1.2.2
Raise -22 to the power of 2.
y=22v±√484v2-4u2⋅(22v)
y=22v±√484v2-4u2⋅(22v)
Step 6.1.3
Factor 4 out of 484v2-4u.
Step 6.1.3.1
Factor 4 out of 484v2.
y=22v±√4(121v2)-4u2⋅(22v)
Step 6.1.3.2
Factor 4 out of -4u.
y=22v±√4(121v2)+4(-u)2⋅(22v)
Step 6.1.3.3
Factor 4 out of 4(121v2)+4(-u).
y=22v±√4(121v2-u)2⋅(22v)
y=22v±√4(121v2-u)2⋅(22v)
Step 6.1.4
Replace all occurrences of u with 22(v⋅(22(vx2)-22(vx)-d)).
y=22v±√4(121v2-(22(v⋅(22(vx2)-22(vx)-d))))2⋅(22v)
Step 6.1.5
Simplify each term.
Step 6.1.5.1
Remove parentheses.
y=22v±√4(121v2-(22(v⋅(22vx2-22vx-d))))2⋅(22v)
Step 6.1.5.2
Apply the distributive property.
y=22v±√4(121v2-(22(v(22vx2)+v(-22vx)+v(-d))))2⋅(22v)
Step 6.1.5.3
Simplify.
Step 6.1.5.3.1
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)+v(-22vx)+v(-d))))2⋅(22v)
Step 6.1.5.3.2
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)+v(-d))))2⋅(22v)
Step 6.1.5.3.3
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)-vd)))2⋅(22v)
Step 6.1.5.4
Simplify each term.
Step 6.1.5.4.1
Multiply v by v by adding the exponents.
Step 6.1.5.4.1.1
Move v.
y=22v±√4(121v2-(22(22(v⋅v)x2-22v(vx)-vd)))2⋅(22v)
Step 6.1.5.4.1.2
Multiply v by v.
y=22v±√4(121v2-(22(22v2x2-22v(vx)-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v(vx)-vd)))2⋅(22v)
Step 6.1.5.4.2
Multiply v by v by adding the exponents.
Step 6.1.5.4.2.1
Move v.
y=22v±√4(121v2-(22(22v2x2-22(v⋅v)x-vd)))2⋅(22v)
Step 6.1.5.4.2.2
Multiply v by v.
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
Step 6.1.5.5
Apply the distributive property.
y=22v±√4(121v2-(22(22v2x2)+22(-22v2x)+22(-vd)))2⋅(22v)
Step 6.1.5.6
Simplify.
Step 6.1.5.6.1
Multiply 22 by 22.
y=22v±√4(121v2-(484(v2x2)+22(-22v2x)+22(-vd)))2⋅(22v)
Step 6.1.5.6.2
Multiply -22 by 22.
y=22v±√4(121v2-(484(v2x2)-484(v2x)+22(-vd)))2⋅(22v)
Step 6.1.5.6.3
Multiply -1 by 22.
y=22v±√4(121v2-(484(v2x2)-484(v2x)-22(vd)))2⋅(22v)
y=22v±√4(121v2-(484(v2x2)-484(v2x)-22(vd)))2⋅(22v)
Step 6.1.5.7
Remove parentheses.
y=22v±√4(121v2-(484v2x2-484v2x-22vd))2⋅(22v)
Step 6.1.5.8
Apply the distributive property.
y=22v±√4(121v2-(484v2x2)-(-484v2x)-(-22vd))2⋅(22v)
Step 6.1.5.9
Simplify.
Step 6.1.5.9.1
Multiply 484 by -1.
y=22v±√4(121v2-484(v2x2)-(-484v2x)-(-22vd))2⋅(22v)
Step 6.1.5.9.2
Multiply -484 by -1.
y=22v±√4(121v2-484(v2x2)+484(v2x)-(-22vd))2⋅(22v)
Step 6.1.5.9.3
Multiply -22 by -1.
y=22v±√4(121v2-484(v2x2)+484(v2x)+22(vd))2⋅(22v)
y=22v±√4(121v2-484(v2x2)+484(v2x)+22(vd))2⋅(22v)
Step 6.1.5.10
Remove parentheses.
y=22v±√4(121v2-484v2x2+484v2x+22vd)2⋅(22v)
y=22v±√4(121v2-484v2x2+484v2x+22vd)2⋅(22v)
Step 6.1.6
Factor 11v out of 121v2-484v2x2+484v2x+22vd.
Step 6.1.6.1
Factor 11v out of 121v2.
y=22v±√4(11v(11v)-484v2x2+484v2x+22vd)2⋅(22v)
Step 6.1.6.2
Factor 11v out of -484v2x2.
y=22v±√4(11v(11v)+11v(-44vx2)+484v2x+22vd)2⋅(22v)
Step 6.1.6.3
Factor 11v out of 484v2x.
y=22v±√4(11v(11v)+11v(-44vx2)+11v(44vx)+22vd)2⋅(22v)
Step 6.1.6.4
Factor 11v out of 22vd.
y=22v±√4(11v(11v)+11v(-44vx2)+11v(44vx)+11v(2d))2⋅(22v)
Step 6.1.6.5
Factor 11v out of 11v(11v)+11v(-44vx2).
y=22v±√4(11v(11v-44vx2)+11v(44vx)+11v(2d))2⋅(22v)
Step 6.1.6.6
Factor 11v out of 11v(11v-44vx2)+11v(44vx).
y=22v±√4(11v(11v-44vx2+44vx)+11v(2d))2⋅(22v)
Step 6.1.6.7
Factor 11v out of 11v(11v-44vx2+44vx)+11v(2d).
y=22v±√4(11v(11v-44vx2+44vx+2d))2⋅(22v)
y=22v±√4⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 6.1.7
Multiply 4 by 11.
y=22v±√44v(11v-44vx2+44vx+2d)2⋅(22v)
Step 6.1.8
Rewrite 44v(11v-44vx2+44vx+2d) as 22⋅(11v(11v-44vx2+44vx+2d)).
Step 6.1.8.1
Factor 4 out of 44.
y=22v±√4(11)v(11v-44vx2+44vx+2d)2⋅(22v)
Step 6.1.8.2
Rewrite 4 as 22.
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 6.1.8.3
Add parentheses.
y=22v±√22⋅(11(v(11v-44vx2+44vx+2d)))2⋅(22v)
Step 6.1.8.4
Add parentheses.
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 6.1.9
Pull terms out from under the radical.
y=22v±2√11v(11v-44vx2+44vx+2d)2⋅(22v)
y=22v±2√11v(11v-44vx2+44vx+2d)2⋅(22v)
Step 6.2
Multiply 2 by 22.
y=22v±2√11v(11v-44vx2+44vx+2d)44v
Step 6.3
Simplify 22v±2√11v(11v-44vx2+44vx+2d)44v.
y=11v±√11v(11v-44vx2+44vx+2d)22v
y=11v±√11v(11v-44vx2+44vx+2d)22v
Step 7
Step 7.1
Simplify the numerator.
Step 7.1.1
Add parentheses.
y=22v±√(-22v)2-4⋅(22(v⋅(22(vx2)-22(vx)-d)))2⋅(22v)
Step 7.1.2
Let u=22(v⋅(22(vx2)-22(vx)-d)). Substitute u for all occurrences of 22(v⋅(22(vx2)-22(vx)-d)).
Step 7.1.2.1
Apply the product rule to -22v.
y=22v±√(-22)2v2-4⋅u2⋅(22v)
Step 7.1.2.2
Raise -22 to the power of 2.
y=22v±√484v2-4u2⋅(22v)
y=22v±√484v2-4u2⋅(22v)
Step 7.1.3
Factor 4 out of 484v2-4u.
Step 7.1.3.1
Factor 4 out of 484v2.
y=22v±√4(121v2)-4u2⋅(22v)
Step 7.1.3.2
Factor 4 out of -4u.
y=22v±√4(121v2)+4(-u)2⋅(22v)
Step 7.1.3.3
Factor 4 out of 4(121v2)+4(-u).
y=22v±√4(121v2-u)2⋅(22v)
y=22v±√4(121v2-u)2⋅(22v)
Step 7.1.4
Replace all occurrences of u with 22(v⋅(22(vx2)-22(vx)-d)).
y=22v±√4(121v2-(22(v⋅(22(vx2)-22(vx)-d))))2⋅(22v)
Step 7.1.5
Simplify each term.
Step 7.1.5.1
Remove parentheses.
y=22v±√4(121v2-(22(v⋅(22vx2-22vx-d))))2⋅(22v)
Step 7.1.5.2
Apply the distributive property.
y=22v±√4(121v2-(22(v(22vx2)+v(-22vx)+v(-d))))2⋅(22v)
Step 7.1.5.3
Simplify.
Step 7.1.5.3.1
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)+v(-22vx)+v(-d))))2⋅(22v)
Step 7.1.5.3.2
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)+v(-d))))2⋅(22v)
Step 7.1.5.3.3
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)-vd)))2⋅(22v)
Step 7.1.5.4
Simplify each term.
Step 7.1.5.4.1
Multiply v by v by adding the exponents.
Step 7.1.5.4.1.1
Move v.
y=22v±√4(121v2-(22(22(v⋅v)x2-22v(vx)-vd)))2⋅(22v)
Step 7.1.5.4.1.2
Multiply v by v.
y=22v±√4(121v2-(22(22v2x2-22v(vx)-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v(vx)-vd)))2⋅(22v)
Step 7.1.5.4.2
Multiply v by v by adding the exponents.
Step 7.1.5.4.2.1
Move v.
y=22v±√4(121v2-(22(22v2x2-22(v⋅v)x-vd)))2⋅(22v)
Step 7.1.5.4.2.2
Multiply v by v.
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
Step 7.1.5.5
Apply the distributive property.
y=22v±√4(121v2-(22(22v2x2)+22(-22v2x)+22(-vd)))2⋅(22v)
Step 7.1.5.6
Simplify.
Step 7.1.5.6.1
Multiply 22 by 22.
y=22v±√4(121v2-(484(v2x2)+22(-22v2x)+22(-vd)))2⋅(22v)
Step 7.1.5.6.2
Multiply -22 by 22.
y=22v±√4(121v2-(484(v2x2)-484(v2x)+22(-vd)))2⋅(22v)
Step 7.1.5.6.3
Multiply -1 by 22.
y=22v±√4(121v2-(484(v2x2)-484(v2x)-22(vd)))2⋅(22v)
y=22v±√4(121v2-(484(v2x2)-484(v2x)-22(vd)))2⋅(22v)
Step 7.1.5.7
Remove parentheses.
y=22v±√4(121v2-(484v2x2-484v2x-22vd))2⋅(22v)
Step 7.1.5.8
Apply the distributive property.
y=22v±√4(121v2-(484v2x2)-(-484v2x)-(-22vd))2⋅(22v)
Step 7.1.5.9
Simplify.
Step 7.1.5.9.1
Multiply 484 by -1.
y=22v±√4(121v2-484(v2x2)-(-484v2x)-(-22vd))2⋅(22v)
Step 7.1.5.9.2
Multiply -484 by -1.
y=22v±√4(121v2-484(v2x2)+484(v2x)-(-22vd))2⋅(22v)
Step 7.1.5.9.3
Multiply -22 by -1.
y=22v±√4(121v2-484(v2x2)+484(v2x)+22(vd))2⋅(22v)
y=22v±√4(121v2-484(v2x2)+484(v2x)+22(vd))2⋅(22v)
Step 7.1.5.10
Remove parentheses.
y=22v±√4(121v2-484v2x2+484v2x+22vd)2⋅(22v)
y=22v±√4(121v2-484v2x2+484v2x+22vd)2⋅(22v)
Step 7.1.6
Factor 11v out of 121v2-484v2x2+484v2x+22vd.
Step 7.1.6.1
Factor 11v out of 121v2.
y=22v±√4(11v(11v)-484v2x2+484v2x+22vd)2⋅(22v)
Step 7.1.6.2
Factor 11v out of -484v2x2.
y=22v±√4(11v(11v)+11v(-44vx2)+484v2x+22vd)2⋅(22v)
Step 7.1.6.3
Factor 11v out of 484v2x.
y=22v±√4(11v(11v)+11v(-44vx2)+11v(44vx)+22vd)2⋅(22v)
Step 7.1.6.4
Factor 11v out of 22vd.
y=22v±√4(11v(11v)+11v(-44vx2)+11v(44vx)+11v(2d))2⋅(22v)
Step 7.1.6.5
Factor 11v out of 11v(11v)+11v(-44vx2).
y=22v±√4(11v(11v-44vx2)+11v(44vx)+11v(2d))2⋅(22v)
Step 7.1.6.6
Factor 11v out of 11v(11v-44vx2)+11v(44vx).
y=22v±√4(11v(11v-44vx2+44vx)+11v(2d))2⋅(22v)
Step 7.1.6.7
Factor 11v out of 11v(11v-44vx2+44vx)+11v(2d).
y=22v±√4(11v(11v-44vx2+44vx+2d))2⋅(22v)
y=22v±√4⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 7.1.7
Multiply 4 by 11.
y=22v±√44v(11v-44vx2+44vx+2d)2⋅(22v)
Step 7.1.8
Rewrite 44v(11v-44vx2+44vx+2d) as 22⋅(11v(11v-44vx2+44vx+2d)).
Step 7.1.8.1
Factor 4 out of 44.
y=22v±√4(11)v(11v-44vx2+44vx+2d)2⋅(22v)
Step 7.1.8.2
Rewrite 4 as 22.
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 7.1.8.3
Add parentheses.
y=22v±√22⋅(11(v(11v-44vx2+44vx+2d)))2⋅(22v)
Step 7.1.8.4
Add parentheses.
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 7.1.9
Pull terms out from under the radical.
y=22v±2√11v(11v-44vx2+44vx+2d)2⋅(22v)
y=22v±2√11v(11v-44vx2+44vx+2d)2⋅(22v)
Step 7.2
Multiply 2 by 22.
y=22v±2√11v(11v-44vx2+44vx+2d)44v
Step 7.3
Simplify 22v±2√11v(11v-44vx2+44vx+2d)44v.
y=11v±√11v(11v-44vx2+44vx+2d)22v
Step 7.4
Change the ± to +.
y=11v+√11v(11v-44vx2+44vx+2d)22v
y=11v+√11v(11v-44vx2+44vx+2d)22v
Step 8
Step 8.1
Simplify the numerator.
Step 8.1.1
Add parentheses.
y=22v±√(-22v)2-4⋅(22(v⋅(22(vx2)-22(vx)-d)))2⋅(22v)
Step 8.1.2
Let u=22(v⋅(22(vx2)-22(vx)-d)). Substitute u for all occurrences of 22(v⋅(22(vx2)-22(vx)-d)).
Step 8.1.2.1
Apply the product rule to -22v.
y=22v±√(-22)2v2-4⋅u2⋅(22v)
Step 8.1.2.2
Raise -22 to the power of 2.
y=22v±√484v2-4u2⋅(22v)
y=22v±√484v2-4u2⋅(22v)
Step 8.1.3
Factor 4 out of 484v2-4u.
Step 8.1.3.1
Factor 4 out of 484v2.
y=22v±√4(121v2)-4u2⋅(22v)
Step 8.1.3.2
Factor 4 out of -4u.
y=22v±√4(121v2)+4(-u)2⋅(22v)
Step 8.1.3.3
Factor 4 out of 4(121v2)+4(-u).
y=22v±√4(121v2-u)2⋅(22v)
y=22v±√4(121v2-u)2⋅(22v)
Step 8.1.4
Replace all occurrences of u with 22(v⋅(22(vx2)-22(vx)-d)).
y=22v±√4(121v2-(22(v⋅(22(vx2)-22(vx)-d))))2⋅(22v)
Step 8.1.5
Simplify each term.
Step 8.1.5.1
Remove parentheses.
y=22v±√4(121v2-(22(v⋅(22vx2-22vx-d))))2⋅(22v)
Step 8.1.5.2
Apply the distributive property.
y=22v±√4(121v2-(22(v(22vx2)+v(-22vx)+v(-d))))2⋅(22v)
Step 8.1.5.3
Simplify.
Step 8.1.5.3.1
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)+v(-22vx)+v(-d))))2⋅(22v)
Step 8.1.5.3.2
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)+v(-d))))2⋅(22v)
Step 8.1.5.3.3
Rewrite using the commutative property of multiplication.
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v(vx2)-22v(vx)-vd)))2⋅(22v)
Step 8.1.5.4
Simplify each term.
Step 8.1.5.4.1
Multiply v by v by adding the exponents.
Step 8.1.5.4.1.1
Move v.
y=22v±√4(121v2-(22(22(v⋅v)x2-22v(vx)-vd)))2⋅(22v)
Step 8.1.5.4.1.2
Multiply v by v.
y=22v±√4(121v2-(22(22v2x2-22v(vx)-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v(vx)-vd)))2⋅(22v)
Step 8.1.5.4.2
Multiply v by v by adding the exponents.
Step 8.1.5.4.2.1
Move v.
y=22v±√4(121v2-(22(22v2x2-22(v⋅v)x-vd)))2⋅(22v)
Step 8.1.5.4.2.2
Multiply v by v.
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
y=22v±√4(121v2-(22(22v2x2-22v2x-vd)))2⋅(22v)
Step 8.1.5.5
Apply the distributive property.
y=22v±√4(121v2-(22(22v2x2)+22(-22v2x)+22(-vd)))2⋅(22v)
Step 8.1.5.6
Simplify.
Step 8.1.5.6.1
Multiply 22 by 22.
y=22v±√4(121v2-(484(v2x2)+22(-22v2x)+22(-vd)))2⋅(22v)
Step 8.1.5.6.2
Multiply -22 by 22.
y=22v±√4(121v2-(484(v2x2)-484(v2x)+22(-vd)))2⋅(22v)
Step 8.1.5.6.3
Multiply -1 by 22.
y=22v±√4(121v2-(484(v2x2)-484(v2x)-22(vd)))2⋅(22v)
y=22v±√4(121v2-(484(v2x2)-484(v2x)-22(vd)))2⋅(22v)
Step 8.1.5.7
Remove parentheses.
y=22v±√4(121v2-(484v2x2-484v2x-22vd))2⋅(22v)
Step 8.1.5.8
Apply the distributive property.
y=22v±√4(121v2-(484v2x2)-(-484v2x)-(-22vd))2⋅(22v)
Step 8.1.5.9
Simplify.
Step 8.1.5.9.1
Multiply 484 by -1.
y=22v±√4(121v2-484(v2x2)-(-484v2x)-(-22vd))2⋅(22v)
Step 8.1.5.9.2
Multiply -484 by -1.
y=22v±√4(121v2-484(v2x2)+484(v2x)-(-22vd))2⋅(22v)
Step 8.1.5.9.3
Multiply -22 by -1.
y=22v±√4(121v2-484(v2x2)+484(v2x)+22(vd))2⋅(22v)
y=22v±√4(121v2-484(v2x2)+484(v2x)+22(vd))2⋅(22v)
Step 8.1.5.10
Remove parentheses.
y=22v±√4(121v2-484v2x2+484v2x+22vd)2⋅(22v)
y=22v±√4(121v2-484v2x2+484v2x+22vd)2⋅(22v)
Step 8.1.6
Factor 11v out of 121v2-484v2x2+484v2x+22vd.
Step 8.1.6.1
Factor 11v out of 121v2.
y=22v±√4(11v(11v)-484v2x2+484v2x+22vd)2⋅(22v)
Step 8.1.6.2
Factor 11v out of -484v2x2.
y=22v±√4(11v(11v)+11v(-44vx2)+484v2x+22vd)2⋅(22v)
Step 8.1.6.3
Factor 11v out of 484v2x.
y=22v±√4(11v(11v)+11v(-44vx2)+11v(44vx)+22vd)2⋅(22v)
Step 8.1.6.4
Factor 11v out of 22vd.
y=22v±√4(11v(11v)+11v(-44vx2)+11v(44vx)+11v(2d))2⋅(22v)
Step 8.1.6.5
Factor 11v out of 11v(11v)+11v(-44vx2).
y=22v±√4(11v(11v-44vx2)+11v(44vx)+11v(2d))2⋅(22v)
Step 8.1.6.6
Factor 11v out of 11v(11v-44vx2)+11v(44vx).
y=22v±√4(11v(11v-44vx2+44vx)+11v(2d))2⋅(22v)
Step 8.1.6.7
Factor 11v out of 11v(11v-44vx2+44vx)+11v(2d).
y=22v±√4(11v(11v-44vx2+44vx+2d))2⋅(22v)
y=22v±√4⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 8.1.7
Multiply 4 by 11.
y=22v±√44v(11v-44vx2+44vx+2d)2⋅(22v)
Step 8.1.8
Rewrite 44v(11v-44vx2+44vx+2d) as 22⋅(11v(11v-44vx2+44vx+2d)).
Step 8.1.8.1
Factor 4 out of 44.
y=22v±√4(11)v(11v-44vx2+44vx+2d)2⋅(22v)
Step 8.1.8.2
Rewrite 4 as 22.
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 8.1.8.3
Add parentheses.
y=22v±√22⋅(11(v(11v-44vx2+44vx+2d)))2⋅(22v)
Step 8.1.8.4
Add parentheses.
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
y=22v±√22⋅(11v(11v-44vx2+44vx+2d))2⋅(22v)
Step 8.1.9
Pull terms out from under the radical.
y=22v±2√11v(11v-44vx2+44vx+2d)2⋅(22v)
y=22v±2√11v(11v-44vx2+44vx+2d)2⋅(22v)
Step 8.2
Multiply 2 by 22.
y=22v±2√11v(11v-44vx2+44vx+2d)44v
Step 8.3
Simplify 22v±2√11v(11v-44vx2+44vx+2d)44v.
y=11v±√11v(11v-44vx2+44vx+2d)22v
Step 8.4
Change the ± to -.
y=11v-√11v(11v-44vx2+44vx+2d)22v
y=11v-√11v(11v-44vx2+44vx+2d)22v
Step 9
The final answer is the combination of both solutions.
y=11v+√11v(11v-44vx2+44vx+2d)22v
y=11v-√11v(11v-44vx2+44vx+2d)22v
Step 10
Set the radicand in √11v(11v-44vx2+44vx+2d) greater than or equal to 0 to find where the expression is defined.
11v(11v-44vx2+44vx+2d)≥0
Step 11
Step 11.1
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
v=0
11v-44vx2+44vx+2d=0
Step 11.2
Set v equal to 0.
v=0
Step 11.3
Set 11v-44vx2+44vx+2d equal to 0 and solve for v.
Step 11.3.1
Set 11v-44vx2+44vx+2d equal to 0.
11v-44vx2+44vx+2d=0
Step 11.3.2
Solve 11v-44vx2+44vx+2d=0 for v.
Step 11.3.2.1
Subtract 2d from both sides of the equation.
11v-44vx2+44vx=-2d
Step 11.3.2.2
Factor 11v out of 11v-44vx2+44vx.
Step 11.3.2.2.1
Factor 11v out of 11v.
11v(1)-44vx2+44vx=-2d
Step 11.3.2.2.2
Factor 11v out of -44vx2.
11v(1)+11v(-4x2)+44vx=-2d
Step 11.3.2.2.3
Factor 11v out of 44vx.
11v(1)+11v(-4x2)+11v(4x)=-2d
Step 11.3.2.2.4
Factor 11v out of 11v(1)+11v(-4x2).
11v(1-4x2)+11v(4x)=-2d
Step 11.3.2.2.5
Factor 11v out of 11v(1-4x2)+11v(4x).
11v(1-4x2+4x)=-2d
11v(1-4x2+4x)=-2d
Step 11.3.2.3
Reorder terms.
11v(-4x2+4x+1)=-2d
Step 11.3.2.4
Divide each term in 11v(-4x2+4x+1)=-2d by 11(-4x2+4x+1) and simplify.
Step 11.3.2.4.1
Divide each term in 11v(-4x2+4x+1)=-2d by 11(-4x2+4x+1).
11v(-4x2+4x+1)11(-4x2+4x+1)=-2d11(-4x2+4x+1)
Step 11.3.2.4.2
Simplify the left side.
Step 11.3.2.4.2.1
Cancel the common factor of 11.
Step 11.3.2.4.2.1.1
Cancel the common factor.
11v(-4x2+4x+1)11(-4x2+4x+1)=-2d11(-4x2+4x+1)
Step 11.3.2.4.2.1.2
Rewrite the expression.
v(-4x2+4x+1)-4x2+4x+1=-2d11(-4x2+4x+1)
v(-4x2+4x+1)-4x2+4x+1=-2d11(-4x2+4x+1)
Step 11.3.2.4.2.2
Cancel the common factor of -4x2+4x+1.
Step 11.3.2.4.2.2.1
Cancel the common factor.
v(-4x2+4x+1)-4x2+4x+1=-2d11(-4x2+4x+1)
Step 11.3.2.4.2.2.2
Divide v by 1.
v=-2d11(-4x2+4x+1)
v=-2d11(-4x2+4x+1)
v=-2d11(-4x2+4x+1)
Step 11.3.2.4.3
Simplify the right side.
Step 11.3.2.4.3.1
Move the negative in front of the fraction.
v=-2d11(-4x2+4x+1)
Step 11.3.2.4.3.2
Factor -1 out of -4x2.
v=-2d11(-(4x2)+4x+1)
Step 11.3.2.4.3.3
Factor -1 out of 4x.
v=-2d11(-(4x2)-(-4x)+1)
Step 11.3.2.4.3.4
Factor -1 out of -(4x2)-(-4x).
v=-2d11(-(4x2-4x)+1)
Step 11.3.2.4.3.5
Rewrite 1 as -1(-1).
v=-2d11(-(4x2-4x)-1(-1))
Step 11.3.2.4.3.6
Factor -1 out of -(4x2-4x)-1(-1).
v=-2d11(-(4x2-4x-1))
Step 11.3.2.4.3.7
Simplify the expression.
Step 11.3.2.4.3.7.1
Rewrite -(4x2-4x-1) as -1(4x2-4x-1).
v=-2d11(-1(4x2-4x-1))
Step 11.3.2.4.3.7.2
Move the negative in front of the fraction.
v=--2d11(4x2-4x-1)
Step 11.3.2.4.3.7.3
Multiply -1 by -1.
v=12d11(4x2-4x-1)
Step 11.3.2.4.3.7.4
Multiply 2d11(4x2-4x-1) by 1.
v=2d11(4x2-4x-1)
v=2d11(4x2-4x-1)
v=2d11(4x2-4x-1)
v=2d11(4x2-4x-1)
v=2d11(4x2-4x-1)
v=2d11(4x2-4x-1)
Step 11.4
The final solution is all the values that make 11v(11v-44vx2+44vx+2d)≥0 true.
v=0
v=2d11(4x2-4x-1)
v=0
v=2d11(4x2-4x-1)
Step 12
Set the denominator in 11v+√11v(11v-44vx2+44vx+2d)22v equal to 0 to find where the expression is undefined.
22v=0
Step 13
Step 13.1
Divide each term in 22v=0 by 22.
22v22=022
Step 13.2
Simplify the left side.
Step 13.2.1
Cancel the common factor of 22.
Step 13.2.1.1
Cancel the common factor.
22v22=022
Step 13.2.1.2
Divide v by 1.
v=022
v=022
v=022
Step 13.3
Simplify the right side.
Step 13.3.1
Divide 0 by 22.
v=0
v=0
v=0
Step 14
The domain is all values of v that make the expression defined.
(No(Minimum),No(Maximum)]
Set-Builder Notation:
{v|No(Minimum)<v≤No(Maximum)}