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3 years ago

HOW DO YOU SOLVE X^2 - 4X - 12 = 0

Mathematics

1 answer:

sleet_krkn [62]3 years ago

4 0

x2−4x−12=0

Step 1: Factor left side of equation.(x+2)(x−6)=0

Step 2: Set factors equal to 0.

x+2=0‌‌ or ‌‌x−6=0

x=−2‌‌ or ‌‌x=6

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What is the supplement to: 101 degrees?

saul85 [17]

Answer:

79 degrees

Step-by-step explanation:

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3 years ago

the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.

Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

$2x^2-3x-7+7=0+7$

$2x^2-3x=7$

Divide both sides by 2

$\frac{2x^2-3x}{2}=\frac{7}{2}$

$x^2-\frac{3x}{2}=\frac{7}{2}$

Add (-3/4)² to both sides

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{65}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{3}{4}=\sqrt{\frac{65}{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{4}$

Add 3/4 to both sides

$x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}$

$x=\frac{\sqrt{65}+3}{4}$

similarly solving

$x-\frac{3}{4}=-\sqrt{\frac{65}{16}}$

$x=\frac{-\sqrt{65}+3}{4}$

So the solution of the equation will have the values of x such as:

$x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}$

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