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3 years ago

7) 4b2 + 8b + 7 = 4 What the answer

Mathematics

2 answers:

Anit [1.1K]3 years ago

8 0

4b² + 8b + 3 = 0

4b² + 2b + 6b + 3 = 0

2b(2b + 1) + 3(2b + 1) = 0

(2b + 3)(2b + 1) = 0

b = -1/2 or b = -3/2

san4es73 [151]3 years ago

7 0

B= -1/2 or b= -3/2

1. Subtract 4 from both sides.
2. Factor left side of equation.
3. Set factors equal to 0.

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What is the value of x in the equation -2/3 x+9= 4/3x-3?

Anvisha [2.4K]

Answer:

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

-2/3*x+9=(4/3*x-3)=?

Step by step solution :

STEP

Simplify —

Equation at the end of step

2 4

((0-(—•x))+9)-((—•x)-3) = 0

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STEP

Rewriting the whole as an Equivalent Fraction

2.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 3 as the denominator :

3 3 • 3

3 = — = —————

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Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

4x - (3 • 3) 4x - 9

———————————— = ——————

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Equation at the end of step

2 (4x - 9)

((0 - (— • x)) + 9) - ———————— = 0

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STEP

Simplify —

Equation at the end of step

2 (4x - 9)

((0 - (— • x)) + 9) - ———————— = 0

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Rewriting the whole as an Equivalent Fraction

4.1 Adding a whole to a fraction

Rewrite the whole as a fraction using 3 as the denominator :

9 9 • 3

9 = — = —————

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Adding fractions that have a common denominator :

4.2 Adding up the two equivalent fractions

-2x + 9 • 3 27 - 2x

——————————— = ———————

3 3

Equation at the end of step

(27 - 2x) (4x - 9)

————————— - ———————— = 0

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Adding fractions which have a common denominator

5.1 Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

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Pulling out like terms

6.1 Pull out like factors :

36 - 6x = -6 • (x - 6)

Equation at the end of step

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When a fraction equals zero

7.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

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Here's how:

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The equation now takes the shape :

-6 • (x-6) = 0

Equations which are never true:

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Solving a Single Variable Equation:

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Vitek1552 [10]

Answer:

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Step-by-step explanation:

Step1:-

a) Given second order homogenous constant co-efficient equation

$y^{ll} - 4y^{l}-12y=0$

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Step 2:-

b) Let f(D) = $(D^{2} -4D-12)y$

Then the auxiliary equation is $(m^{2} -4m-12)=0$

Find the factors of the auxiliary equation is

$m^{2} -6m+2m-12=0$

m(m-6) + 2(m-6) =0

m+2 =0 and m-6=0

m=-2 and m=6

The roots are real and different

The general solution $y = c_{1} e^{-m_{1} x} + c_{2} e^{m_{2} x}$

the roots are $m_{1} = -2 and m_{2} = 6$

The general solution of given differential equation is

$y = c_{1} e^{-2x} + c_{2} e^{6 x}$

Step 3:-

C) Given initial conditions are y(0) =1 and y1 (0) =-26

The general solution of given differential equation is

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$ .....(1)

substitute x =0 and y(0) =1

$y(0) = c_{1} e^{0} + c_{2} e^{0}$

$1 = c_{1} + c_{2}$ .........(2)

Differentiating equation (1) with respective to 'x'

$y^l(x) = -2c_{1} e^{-2x} + 6c_{2} e^{6 x}$

substitute x= o and y1 (0) =-26

$-26 = -2c_{1} e^{0} + 6c_{2} e^{0}$

$-2c_{1} + 6c_{2} = -26$ .............(3)

solving (2) and (3) by using substitution method

$substitute c_{2} =1- c_{1}$ in equation (3)

$-2c_{1} + 6(1-c_{1}) = -26$

on simplification , we get

$-2c_{1} + 6(1)-6c_{1}) = -26$

$-8c_{1} = -32$

dividing by'8' we get $c_{1} =4$

substitute $c_{1} =4$ in equation $1 = c_{1} + c_{2}$

so $c_{2} = 1-4 = -3$

now substitute $c_{1} =4 and c_{2} =-3$ in general solution

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$

$y(x) = 4 e^{-2x} -3 e^{6 x}$

now the initial value problem

$y(x) = 4 e^{-2x} -3 e^{6 x}$

7 0

4 years ago