What is the solution to x2 + 10x + 28 = 0 when written in the form a ± bi?

2 answers:

8 0

$x^2+10x+28=0\\\\a=1;\ b=10;\ c=28\\\\\Delta=b^2-4ac\\\\\Delta=10^2-4\cdot1\cdot28=100-112=-112\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{-12}=i\sqrt{12}=i\sqrt{4\cdot3}=i\sqrt4\cdot\sqrt3=2i\sqrt3$

$x_1=\dfrac{-10-2i\sqrt3}{2\cdot1}=-5-i\sqrt3\\\\x_2=\dfrac{-10+2i\sqrt3}{2\cdot1}=-5+i\sqrt3$

mr Goodwill [35]2 years ago

4 0

Use the quadratic formular:
x=[-10+√(10²-4*1*28)]/2*1 or x=[-10-√(10²-4*1*28)]/2*1
x=[-10+√(-12)]/2 or x=[-10-√(-12)]/2
x=-5+√3i or x=-5-√3i

You might be interested in

What is mѕТ? ооо 0100° o 108° o 130° 0 160°

lara [203]

Answer:

most is 160 hope I helped

8 0

2 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$