Factorising quadratics All questions please to be completed. I’m stuck.

Mathematics

1 answer:

Alex777 [14]3 years ago

5 0

Answer:

1. (x +1)(x - 1)

2. (x+5)(x-5)

3. (x+12)(x-12)

4. (x+14)(x-14)

5. (x+5)(x-7)

Step-by-step explanation:

The first 4 expressions can be factored with the difference of squares type of factoring.

For #5, the odd one out is (x+5)(x-7) because it isn't a factor of any of the standard form equations, while the others are

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You roll a single die numbered from 1 to 6. What are the odds of rolling a 3

vodomira [7]

Answer:

1/6

Step-by-step explanation:

there are 6 numbers on a die, and only one of them is 3.

so you would get 1/6 (because only one number out of 6 is 3)

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3 years ago

What are the solutions of this quadratic equation?X2 - 10x= -34A.r=-8, -2B.r= 5 + 3iC.r=-5 + 3iD.r=-5 + 159

ololo11 [35]

The given equation is-

$x^2-10x=-34$

First, we move the independent term to the other side.

$x^2-10x+34=0$

Now, we have to use the quadratic equation to find the solutions.-

$x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where, a = 1, b = -10, and c = 34.

Replacing these values in the formula, we have.

$\begin{gathered} x_{1,2}=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(1)(34)}}{2(1)} \\ x_{1,2}=\frac{10\pm\sqrt[]{100-136}}{2}=\frac{10\pm\sqrt[]{-36}}{2} \end{gathered}$

But, there's no square root of -36 because it's a negative. To solve this issue, we use complex numbers that way, we would have solutions.

$x_{1,2}=\frac{10\pm\sqrt[]{36}i}{2}=\frac{10\pm6i}{2}=5\pm3i$ <h2>Therefore, the solutions are</h2> $\begin{gathered} x_1=5+3i \\ x_2=5-3i \end{gathered}$ The right answer is B.