Question

write a quadratic polynomial sum of whose zeroes is -3 and product is 2
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Answer 1

Respuesta:

$x^{2}+3x+2=0$

Explicación paso a paso:

When they tell us "zeroes" they're talking about the solutions of the quadratic equation (or those values that makes the polynom results in zero), knowing that, we can tell:

$x_{1}+ x_{2} =-3$ "write a quadratic polynomial of whose zeroes is -3"

$x_{1} x_{2} =2$  "and product is 2"

Where X1 and X2 are two of possible solutions to the quadratic equation.

So:

$\left \{ {{x_{1}+ x_{2} =-3} \atop {x_{1} x_{2} =2}} \right.$

Taking the second equation:

$x_{1} x_{2} =2\\x_{1}=\frac{2}{x_{2}}$

Replacing in first equation:

$x_{1}+ x_{2} =-3\\\\\frac{2}{x_{2}} + x_{2} =-3$  Multiply by X2 every term of the equation (to make it easier)

$\frac{2}{x_{2}}(x_{2}) + x_{2} (x_{2})=-3(x_{2})\\\\2+(x_{2})^{2}=-3x_{2}\\\\(x_{2})^{2}+3x_{2}+2=0$  Now, we need to look for two numbers that their product is +2 and their sum is + 3, those numbers are +1 and +2, so:

To make the resolution easier to read I will replace X2 for K:

$k^{2} +3k+2=0\\(k+2)(k+1)=0\\\\k_{1}=-2\\k_{2}=-1$

We can take any of those two solutions, that's because doesn't matter what solution you take, you will get the desired polynom.

For this example I will take K1 as X2 = -2

Then, in the second equation:

$x_{1} x_{2} =2\\\\x_{1} (-2)=2\\\\x_{1}=\frac{2}{-2} \\\\x_{1}=-1$

Now we have both solutions of the quadratic polynom.

$x_{1}=-1\\x_{1}+1=0\\\\x_{2}=-2\\x_{2}+2=0$

Where their sum is: -1 - 2 = -3

And their product is: (-1)(-2) = 2

Knowing this, we can finish:

$(x+1)(x+2)=0\\x^{2} +3x+2=0$

Hope it helps!