Solve the following:

1.) x2 + 2x + 9 = 0

2.) x2 - 7x = -13

3.) -3x2 - 4x - 4 = 0

4.) 3x2 = -12x - 15

5.) -2x2 - 16x - 44 = 0

Skip to content# Solve the following: 1.) x2 + 2x + 9 = 0 2.) x2 – 7x = -13 3.) -3×2 – 4x – 4 = 0 4.) 3×2 = -12x – 15

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Solve the following:

1.) x2 + 2x + 9 = 0

2.) x2 - 7x = -13

3.) -3x2 - 4x - 4 = 0

4.) 3x2 = -12x - 15

5.) -2x2 - 16x - 44 = 0

4x4956

Step-by-step explanation:

thats what u do

we are to solve each trinomial. this should cost more points than any other items to be solved.

The discriminants of all equations are negative hence the roots are imaginary numbers

1. x = -1 +- 2.8284 i 2. x = 3.5 +- 0.8660 i 3. x = -2/3 +- 0.9428 i 4. x = -2 +- i 5. x = -4 +- 2.4495 i

[tex]x^{2}+2x+9=0 \\ \\ x^{2}+2x+1+8=0 \\ \\ (x+1)^{2}+(2 \sqrt{2})^{2}=0 \\ \\ (x+1+2 \sqrt{2}i)(x+1-2 \sqrt{2}i)=0 \\ \\ x+1+2 \sqrt{2}i=0 \ \vee \ x+1-2 \sqrt{2}i=0 \\ \\ x=-1-2 \sqrt{2}i \ \vee \ x=-1+2 \sqrt{2}i[/tex]

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[tex]a^{2}+b^{2}=(a+bi)(a-bi)[/tex]

x has no real solution

Step-by-step explanation:

Our equation is qudratic equation so the method we will follow to solve it is using the dicriminant :

Let Δ be the dicriminant a=1b=2c=9 Δ= 2²-4*1*9 =4-36=-32 we notice that Δ≤0⇒x has no real solution

the anwser is b

Step-by-step explanation:

572727 go to school son

Hi there!

We can use the discriminant of the quadratic formula to find the number of real solutions to this (or any other quadratic) equation.

The discriminant can be found by plugging in the data from the equation into the following formula:

[tex]d = {b}^{2} - 4ac \\ with \: y \: = a {x}^{2} + bx + c[/tex]

When we fill in the data from the question we get

[tex]d = ( - 2) {}^{2} - 4 \times 1 \times 9 = 4 - 36 = - 32[/tex]

Since d < 0 the equation has zero real solutions.

x = 1 + [tex]\frac{\sqrt{32} }{2}[/tex]i or 1 -[tex]\frac{\sqrt{32} }{2}[/tex]i , therefore no real number solution.

Step-by-step explanation:

x² - 2x + 9 = 0

We are going to use formula method to find the solution to the equation below

x = -b ± √ b² - 4ac / 2a

From the equation given;

a = 1 b=-2 and c = 9

We can now proceed to insert the values into the formula;

x = -b ± √ b² - 4ac / 2a

x = 2 ± √ -2² - 4(1)(9) / 2(1)

x = 2 ± √ 4 - 36 / 2

x = 2 ±√ -32 / 2

x = [tex]\frac{2}{2}[/tex] ± [tex]\frac{\sqrt{-32} }{2}[/tex]

x = 1 ± [tex]\frac{\sqrt{32} }{2} \sqrt{-1}[/tex]

x = 1 ±[tex]\frac{\sqrt{32} }{2}[/tex]i

Either x = 1 + [tex]\frac{\sqrt{32} }{2}[/tex]i or 1 -[tex]\frac{\sqrt{32} }{2}[/tex]i

Therefore no real number solutions to the equation

Option 1 x² + 11x +30 = 0 is correct.

Step-by-step explanation:

Zero product property states that if ab=0 then a=0 and b =0

The equation 1 can be factored easily i.e,

x² + 11x +30 = 0

x² + 5x+6x +30 = 0

x(x+5)+6(x+5)=0

(x+6)(x+5)=0

so, applying zero product property

(x+6)=0 and (x+5)=0

x=-6 and x=-5

The rest of the options 2,3 and 4 are prime equations, and they could be solved using quadratic formula or by completing the square but not using zero product property.

So, Option 1 is correct.

[tex]x^2+2x+9=0 \\ \\ a=1 \\ b=2 \\ c=9 \\ b^2-4ac=2^2-4 \times 1 \times 9=4-36=-32 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-2 \pm \sqrt{-32}}{2 \times 1}=\frac{-2 \pm \sqrt{-16 \times 2}}{2}=\frac{-2 \pm 4i\sqrt{2}}{2}=-1 \pm 2i\sqrt{2}[/tex]

The answer is D.