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Hallie Gutmann

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βˆ™ 3y ago
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βˆ™ 12y ago

Since we don't know how many numbers multiply to 4 and the same on addition give -6.

Let us say there are n variables which are:

a1, a2, a3,..., an.

According to question:

a1 x a2 x a3 x... x an = 4

a1 + a2 + a3 + ... + an = -6

We obtain two equations.

Solution to the problem can be obtained only if the number of variables = number of equations.

Since we have only two equations so the number of variables should be 2.

So we can answer to the problem we have these two equations:

a1 x a2 = 4.....(1)

a1 + a2 = -6....(2)

From equation 2 we have a1 = -6 - a2

Putting a1 = -6 - a2 in equation we get:

(-6 - a2) x a2 = 4

-6a2 - a22 = 4

a22 + 6a2 + 4 = 0 which is a quadratic equation in one variable.

General form is: px2 + qx + r = 0

And its solution is: x = [-q ± (q2 - 4pr)1/2]/2p

On comparing we get p = 1, q = 6 and r = 4

And solution is: a2 = [-6 ± (62 - 4x1x4)1/2]/2 = -3 ±√5

When a2 = -3 - √5 then a1 = - 6 - (-3 - √5) = -3 + √5

When a2= -3 + √5 then a1 = -6 - (-3 + √5) = -3 - √5

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Q: What multiplies to be 4 but adds to be negative 6?
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