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enyata [817]
3 years ago
10

One zero of x3 – 3x2 – 6x + 8 = 0 is -2. What are the other zeros of the function?

Mathematics
1 answer:
hram777 [196]3 years ago
7 0
Consider this solution, if it is possible, check the result:
1) if to divide polynomial 'x³-3x²-6x+8' on 'x+2' result is: x²-5x+4, then x³-3x²-6x+8=(x+2)(x²-5x+4);
2) x²-5x+4=(x-1)(x-4), then
3) x³-3x²-6x+8=(x+2)(x-1)(x-4).
4) zeros are: -2;1;4. Other zeros are: 1 and 4.
Answer: 1;4.
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The answer is "The discriminant is positive"

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When the discriminant is positive there is two real solutions.

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Please help!!
ZanzabumX [31]

The two linear equations in two variable is:

12 x + 3 y = 40

7 x - 4 y = 38

(a) For a system of equations in two Variable

a x + by = c

p x + q y = r

It will have unique solution , when

\frac{a}{p}\neq \frac{b}{q}\neq\frac{c}{r}

As, you can see  that in the two equation Provided above

\frac{12}{7}\neq \frac{3}{-4}\neq \frac{40}{38}

So, we can say the system of equation given here has unique solution.

(b). If point (2.5, -3.4) satisfies both the equations, then it will be solution of the system of equation, otherwise not.

1. 12 x+3 y=40

2. 7 x-4 y=38

Substituting , x= 2.5 , and y= -3.4 in equation (1) and (2),

L.H.S of Equation (1)= 1 2 × 2.5 + 3 × (-3.4)

                             = 30 -10.20

                               = 19.80≠ R.H.S that is 40.

Similarly, L H S of equation (2)= 7 × (2.5) - 4 × (-3.4)

                                                  = 17.5 +13.6

                                                  = 31.1≠R HS that is 38

So, you can Write with 100 % confidence that point (2.5, -3.4) is not a solution of  this system of the equation.


5 0
2 years ago
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