Solve: x+y+ 4z = 6, 3x +2y +z =4 2x+ 2y + z =9 If the roots of the equation ax^2 + bx + c are 6 and 5 what are values of a and b?

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

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We have to solve the system of equations:

x + y + 4z = 6… (1)

3x +2y +z =4… (2)

2x+ 2y + z =9… (3)

Now (2)-(3)

=> 3x +2y +z - 2x- 2y – z = 4-9

=> x = -5

Substituting x= -5 in (1) and (2) we get the two equations:

y + 4z = 11… (4)

2y +z = 19… (5)

5-2*(4)

=> 2y +z – 2y -8z = 19 -22

=> -7z =-3

=> z = 3/7

Substituting z = 3/7 in 2y +z = 19

=> 2y = 19 – 3/7

=> 2y = 130/7

=> y=65/7

Therefore x = -5, y= 65/ 7 and z = 3/7

 

We have to find the values of a, b and c for the quadratic equation ax^2 + bx + c if the roots of the equation are 6 and 5. From the information given we can write

(x-6)(x-5)= ax^2 + bx + c

=> x^2 – 6x -5x +30 = ax^2 + bx + c

=> x^2 -11x +30 = ax^2 + bx + c

Equating the coefficients of x^2 , x and the numeric term for x^2 -11x +30 and ax^2 + bx + c we get

a = 1, b = -11 and c = 30.

Therefore the value for a and b are 1 and -11, respectively

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the given equations of the system are all linear equations, we'll solve the system using another method.

We'll calculate the determinent of the system. The determinant of the system is formed from the coefficients of the variables, x, y and z.

We'll note the determinant as det A.

             1   1   4

det A =  3   2   1

              2   2   1

We'll calculate det A:

det A = 1*2*1 + 3*2*4  + 1*1*2 - 4*2*2 - 2*1*1 - 3*1*1

det A = 2 + 24 + 2 - 16 - 2 - 3

We'll eliminate and combine like terms:

det A = 7

Now, we'll calculate the variable x using Cramer formula:

x = detX / detA

Det X is the determinant whose column of coefficients of the variable that has to be found (in this case x) is substituted by the column of the terms from the right side of the equal (6 , 4 , 9).

            6   1   4

det X = 4   2   1

            9    2   1

det X = 6*2*1 + 4*2*4 + 1*1*9 - 4*2*9 - 6*2*1 - 4*1*1

We'll eliminate like terms:

detX = 32 + 9 - 72 - 12 - 4

det X = -47

x = detX/detA

x = -47/7

            1   6   4

det Y = 3   4   1

             2   9   1

det Y = 4 + 108 + 12 - 32 - 9 - 18

y = detY/detA

y = 65/7

We'll calculate z substituting the values of x and y into the first equation:

x+y+ 4z = 6

4z = 6 - x - y

z = (6  -x - y)/4

z = (6 + 47/7 - 65/7)/4

z = -12/7*4

z = -3/7

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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If x1 and x2 are the roots of an equation, the  (x-x1)(x-x2) = 0  or a(x-x1)(x-x2) = 0 is the required equation.

given roots are x1 = 6 and x2 = 5.

Therefore ax^2+bx+c = 0  or x^2+(b/a)x+c/a = 0 could be wrtten as (x-6)(x-5) = 0

Therefore = x^2+(b/a)x+c/a = x^2-(6+5)x+(-6)(-5)

x^2+(b/a)x+c/a = ax^2 -11x +30.

Equating the coefficients, b/a = -11. and c/a = 30

Thetre for  any a, (other than zero) b = -11a and c = 30a.

To be simple form,  a= 1, b = -11 and c = 30. Or

ax^2 +bx+c = a(x^2+11x+30) for any a except  zero Or

ax^2+bx+c = x^2-11x+30 in a simple form.

a= a, b = 11 and c =

To solve

x+y+4z =6..........(1)

3x+2y+z =4.........(2)

2x+2y+z = 9........(3)

Eq(2)-eq(3) gives: x  = 4-9 = -5. Using this we rewrite  equations (1) and (2):

-5+y+z = 6 Or

y+z = 11...........(3) and

3(-5)+2y+z= 4 or

2y+z = 19................(4)

Eq(4)-eq(3) Gives:

y = 19-11 = 8.

Substituting  x = -5, and y = 3 in (1), we get:

-5+8+z =  6

z = 6+5-8 = 3.

Therefore x = -5,  y= 8 and z = 3.  

 

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