solve the equation by making an appropriate substitution?

Question

Solve the equation by making an appropriate substitution:

(x2 - 3x)2 - 14(x2 - 3x) + 40 = 0

a. {-2, -1, 5, 4}
b. {-2, -1, 10, 4, 5, 4}
c. {10, 4}
d. {5, 4}

Answer 1

do you mean (x2 - 3x)2
or (x^2 - 3x)2
or (x^2 - 3x)^2?
if you mean (x^2 - 3x)^2 - 14(x^2 - 3x) + 40
then let n = (x^2 - 3x)
then you have n^2 - 14n + 40 = 0
or n = 10, n = 4
so x^2 - 3x = 10 or x^2 - 3x = 4
x^2 - 3x - 10 = 0
(x-5)(x+2)=0
x=5 or -2

x^2 - 3x-4=-
(x-4)(x+1)
x=4 or -1
a.

Answer 2

Hai charliemo,

(x^2 - 3x)^2 - 14(x^2 - 3x) + 40 = 0 .....(1)
Let x^2 - 3x = y .....(2)

Then. eqn. (1) can be written as y^2 - 14y + 40 = 0

Factorizing we get, y^2 - 10y -4y + 40 = y(y-10)-4(y-10) = 0
=> (y-10)(y-4) = 0
=> y = 10 or 4

Substituting for y in eqn. (2), we get,
x^2 - 3x = 10 or 4

So we get,
x^2 - 3x -10 = 0 and x^2 - 3x - 4 = 0
=> x^2 - 5x + 2x - 10 = 0 and x^2 - 4x + x - 4 = 0
=> x(x-5)+2(x-5)=0 and x(x-4)+1(x-4)=0
=> (x-5)(x+2)=0 and (x-4)(x+1)=0
=> x=5 or -2 and x= 4 or -1

So, the answer is: a. {-2, -1, 5, 4}

Answer 3

Let y= x^2-3x
Then y^2 -14y +40 = 0
(y-10)(y-4)= 0
So y = 10 or 4
x^2-3x =10
x^2 -3x -10 = 0
(x-5)(x+2) = 0
x = 5 and x= -2
x^2-3x -4 = 0
(x-4)(x+1) = 0
x = 4 and x = -1
So answer is a. {-2, -1, 5, 4}

Answer 4

y=x^2 -3x

y^2 -14y +40 =0

solve the quadratic:

(y-4)(y-10) = 0

y= 4,10

now you can set each number equal to the first equation that I created:

4= x^2 -3x
x^2 -3x -4 =0
(x-4)(x+1) =0
x=4,-1

10 = x^2 -3x
x^2 -3x -10=0
(x-5)(x+2) =0
x=5,-2

now that means that x can equal: 5,-2,4,-1
now you should check each to make sure it works out...

(5^2 -15)^2 -14(5^2-15) +40 =0
100 - 140 +40 =0
0=0

(4^2 -12)^2 -14(4^2-12) +40 =0
16 - 224 +40 = 0
-168 =0
NOT A solution

([-2]^2 +6)^2 -14(5^2 +6) +40 =0
100 - 140 +40 = 0
0=0

([-1]^2 -3)^2 -14(5^2 +3) +40 =0
4 + 28 +40 =0
72 = 0
NOT a solution

so the solutions are 5 and -2

Answer 5

There is NO solution, this is nonsensical, because the original equation has only ONE variable, "x". And what does x2 mean? That is NOT 2 times x, and is not normal syntax.

a, b, c, and d have different numbers of variables; thus the question is absurd.

Laurie

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Answer 6

something is not right with the equation.
i tried for about 5 mnts and i couldnt come up with a solution.
sorry.
i came close 2 A, but it wasnt certain.