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Nana76 [90]
3 years ago
9

Solve x2 − 12x + 5 = 0 using the completing-the-square method. x = six plus or minus the square root of five x = negative six pl

us or minus the square root of five x = six plus or minus the square root of thirty one x = negative six plus or minus the square root of thirty one
Mathematics
1 answer:
Andru [333]3 years ago
5 0

we are given

x^2-12x+5=0

we have to solve it by completing square method

step-1: Move 5 on right side

x^2-12x+5-5=0-5

x^2-12x=-5

step-2: Break middle term

x^2-2*6*x=-5

step-3: Add 6^2 both sides

x^2-2*6*x+(6)^2=-5+(6)^2

(x-6)^2=31

step-3: Solve for x

\sqrt{(x-6)^2} =\sqrt{31}

we wil get two values

First value is

x-6=\sqrt{31}

add both sides 6

x-6+6=6+\sqrt{31}

x=6+\sqrt{31}

Second value is

x-6=-\sqrt{31}

add both sides 6

x-6+6=6-\sqrt{31}

x=6-\sqrt{31}

so, solutions are

x=6+\sqrt{31}

x=6-\sqrt{31}.............Answer

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Contact [7]

2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0

Divide both sides by x^2\,\mathrm dx to get

2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}

Substitute v(x)=\dfrac{y(x)}x, so that \dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}. Then

x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}

The remaining ODE is separable. Separating the variables gives

\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x

Integrate both sides. On the left, split up the integrand into partial fractions.

\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}

\implies v^2+2v+1=a(v^2+1)+(bv+c)v

\implies v^2+2v+1=(a+b)v^2+cv+a

\implies a=1,b=0,c=2

Then

\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v

On the right, we have

\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C

Solving for v(x) explicitly is unlikely to succeed, so we leave the solution in implicit form,

\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C

and finally solve in terms of y(x) by replacing v(x)=\dfrac{y(x)}x:

\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}

7 0
3 years ago
What is X squared +3x-70
Aleksandr-060686 [28]

Answer:

x = 7, x = -10

Step-by-step explanation:

x^2+3x-70 = 0

Use the quadratic formula.

x = \frac{-3 + \sqrt{3^2-4(1)(-70)} }{2(1)} \\x = \frac{-3 - \sqrt{3^2-4(1)(-70)} }{2(1)} \\

Solve.

x = 7, x = -10

You can also factor if you want - that is a faster method.

8 0
3 years ago
Which of the following is true for f(x) = -2 sin(x) - 3?
Naddik [55]

Answer:

Step-by-step explanation:

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Is (x+4)^2equivalent to 2x^2+8x+8
Mashcka [7]

Answer:

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Step-by-step explanation:

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