Given that a and 2a are the roots of the equation 8x^2 + kx + 9 = 0, find the value of a and k. Please help with full solutions )):?
回答 (10)
2016-06-25 4:52 pm
a and 2a are the roots of the equation 8x² + kx + 9 = 0
Product of the two roots :
a * (2a) = 9/8
2a² = 9/8
16a² - 9 = 0
(4a)² - 3² = 0
(4a - 3)(4a + 3) = 0
a = 3/4 or a = -3/4
When a = 3/4 :
Sum of the two roots :
a + (2a) = -k/8
(3/4) + (6/4) = -k/8
k = -18
When a = -3/4 :
Sum of the two roots :
a + (2a) = -k/8
(-3/4) + (-6/4) = -k/8
k = 18
Hence, (a, k) = (3/4, -18) or (-3/4, 18)
Product of the two roots :
a * (2a) = 9/8
2a² = 9/8
16a² - 9 = 0
(4a)² - 3² = 0
(4a - 3)(4a + 3) = 0
a = 3/4 or a = -3/4
When a = 3/4 :
Sum of the two roots :
a + (2a) = -k/8
(3/4) + (6/4) = -k/8
k = -18
When a = -3/4 :
Sum of the two roots :
a + (2a) = -k/8
(-3/4) + (-6/4) = -k/8
k = 18
Hence, (a, k) = (3/4, -18) or (-3/4, 18)
2016-06-25 3:03 pm
Since a and 2a are roots, you should know how the quadratic factors:
8x^2 + kx + 9 = 8(x-a)(x-2a)
8x^2 + kx + 9 = 8x^2 - 24ax + 16a^2
Comparing the coefficients:
k = -24a
9 = 16a^2
Now solve for both a and k.
8x^2 + kx + 9 = 8(x-a)(x-2a)
8x^2 + kx + 9 = 8x^2 - 24ax + 16a^2
Comparing the coefficients:
k = -24a
9 = 16a^2
Now solve for both a and k.
2016-06-25 3:10 pm
Sum of roots:
a + 2a = -k/8
24a + k = 0
Product of roots:
a(2a) = 9/8
a² = 9/16
a + 2a = -k/8
24a + k = 0
Product of roots:
a(2a) = 9/8
a² = 9/16
2016-07-03 1:15 pm
If a and 2a are roots than (x-a) and (x-2a) have to be factors:
8(x-a)(x-2a) = 8(x^2-3ax+2a^2) = 8x^2 -24ax +16a^2 = 8x^2 + kx + 9 = 0
-24a = k and 16a^2 = 9
a = +/- 3/4
if a = 3/4, k = 3/4(-24) = -18
if a = -3/4, k = 3/4(24) = 18
(a,k) = (3/4, -18) or (-3/4, 18)
8(x-a)(x-2a) = 8(x^2-3ax+2a^2) = 8x^2 -24ax +16a^2 = 8x^2 + kx + 9 = 0
-24a = k and 16a^2 = 9
a = +/- 3/4
if a = 3/4, k = 3/4(-24) = -18
if a = -3/4, k = 3/4(24) = 18
(a,k) = (3/4, -18) or (-3/4, 18)
2016-06-30 12:57 pm
Substitute x=a and x=2a to get two simultaneous eqns in a and k :
8a^2+ka+9=0 ...(i)
32a^2+2ka+9=0 ...(ii)
Take (ii) - 2*(i) :
16a^2-9=0
a^2=9/16
a= +/- 3/4.
Take 4*(i) - (ii) :
2ka+27=0
k= -27/2a = -/+ 18.
Answer : (a,k) = (+3/4, -18) or (-3/4, +18).
8a^2+ka+9=0 ...(i)
32a^2+2ka+9=0 ...(ii)
Take (ii) - 2*(i) :
16a^2-9=0
a^2=9/16
a= +/- 3/4.
Take 4*(i) - (ii) :
2ka+27=0
k= -27/2a = -/+ 18.
Answer : (a,k) = (+3/4, -18) or (-3/4, +18).
2016-06-30 1:18 am
Given that a and 2a are the roots of the equation 8x^2 + kx + 9 = 0, .................... [1]
So,
x^2 - 3ax +2a^2 = 0, ........................................ [2]
Equating co-efficients of [1] & [2],
k = -3a, and 2a^2 = 9. ==> a^2 = 4.5, ==> a = sqrt(4.5) = 2.1213,
and
k = -3(2.1213) = -6.3634.
So,
x^2 - 3ax +2a^2 = 0, ........................................ [2]
Equating co-efficients of [1] & [2],
k = -3a, and 2a^2 = 9. ==> a^2 = 4.5, ==> a = sqrt(4.5) = 2.1213,
and
k = -3(2.1213) = -6.3634.
2016-06-29 5:35 pm
8x² + kx + 9 = 0
If "a" is a root, you can factorize (x - a). If "2a" is a root, you can factorize (x - 2a). Then it gives us:
= 8.(x - a).(x - 2a)
= 8.(x² - 2ax - ax + 2a²)
= 8.(x² - 3ax + 2a²)
= 8x² - 24ax + 16a² → then you compare to the polynomial: 8x² + kx + 9
9 = 16a² → a² = 9/16 → a = ± 3/4
k = - 24a → k = - 24.(± 3/4) → k = ± 18
If "a" is a root, you can factorize (x - a). If "2a" is a root, you can factorize (x - 2a). Then it gives us:
= 8.(x - a).(x - 2a)
= 8.(x² - 2ax - ax + 2a²)
= 8.(x² - 3ax + 2a²)
= 8x² - 24ax + 16a² → then you compare to the polynomial: 8x² + kx + 9
9 = 16a² → a² = 9/16 → a = ± 3/4
k = - 24a → k = - 24.(± 3/4) → k = ± 18
2016-06-29 2:19 pm
Put f(x) = 8x^2+kx+9..[1]. a & 2a are zeros of f(x). Then (x-a)(x-2a) is a factor of f(x), ie., x^2-3ax+2a^2
= g(x), say, is a factor of f(x). Then f(x) = 8g(x) because coefficient of x^2 in f(x) is 8. So we have
f(x) = 8x^2 - 24ax + 16a^2..[2]. Equating coefficients of like powers of x in [1] & [2] gives k=-24a..[3]
and 16a^2=9..[4]. Now [4]--> a = (+/-)(3/4). Then k correspondingly = (-/+18). Then values of a & k
are (a , k) = ([3/4], -18) or ([-3/4], 18).
= g(x), say, is a factor of f(x). Then f(x) = 8g(x) because coefficient of x^2 in f(x) is 8. So we have
f(x) = 8x^2 - 24ax + 16a^2..[2]. Equating coefficients of like powers of x in [1] & [2] gives k=-24a..[3]
and 16a^2=9..[4]. Now [4]--> a = (+/-)(3/4). Then k correspondingly = (-/+18). Then values of a & k
are (a , k) = ([3/4], -18) or ([-3/4], 18).
2016-06-27 12:26 pm
8x^2 + kx + 9 = 0,
x=a and 2a are solutions. a should satisfy the equation by putting x=a
8a^2 + ka +9=0,
Similarly, x= 2 a
8(4a^2) +2ka +9=0
multiply first equation by 4. you get
32 a^2 + 4ka +36=0
32 a^2 + 2 ka +9=0
Subtract second equation from the first equation.
2 ka + 27=0
Put back in the first equation
2 a^ 2- 27/2+9=0
2 a^2 = 27/2 -9= 9/2
a^2 = 9/4, a = +/- 3/2,
2 k 3/2=-27
k= -9. The other value k=9
x=a and 2a are solutions. a should satisfy the equation by putting x=a
8a^2 + ka +9=0,
Similarly, x= 2 a
8(4a^2) +2ka +9=0
multiply first equation by 4. you get
32 a^2 + 4ka +36=0
32 a^2 + 2 ka +9=0
Subtract second equation from the first equation.
2 ka + 27=0
Put back in the first equation
2 a^ 2- 27/2+9=0
2 a^2 = 27/2 -9= 9/2
a^2 = 9/4, a = +/- 3/2,
2 k 3/2=-27
k= -9. The other value k=9
2016-06-25 3:47 pm
x = a
x - a = 0
x = 2a
x - 2a = 0
(x - a)(x - 2a) = 0
x^2 - 2ax - ax + 2a^2 = 0
x^2 - 3ax + 2a^2 = 0
8(x^2 - 3ax + 2a^2) = 0
8x^2 - 24ax + 16a^2 = 0
8x^2 - 24ax + 16a^2 = 8x^2 + kx + 9
-24ax = kx
-24a = k
16a^2 = 9
a^2 = 9/16
a = 3/4, -3/4
k = -18, 18
8x^2 - 18x + 9 = 0
8x^2 + 18x + 9 = 0
x - a = 0
x = 2a
x - 2a = 0
(x - a)(x - 2a) = 0
x^2 - 2ax - ax + 2a^2 = 0
x^2 - 3ax + 2a^2 = 0
8(x^2 - 3ax + 2a^2) = 0
8x^2 - 24ax + 16a^2 = 0
8x^2 - 24ax + 16a^2 = 8x^2 + kx + 9
-24ax = kx
-24a = k
16a^2 = 9
a^2 = 9/16
a = 3/4, -3/4
k = -18, 18
8x^2 - 18x + 9 = 0
8x^2 + 18x + 9 = 0
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