Question 268757: How do you solve (2x-7)(x+1)=-4?
also -12=2(x+4)^2?
Answer by persian52(161)   (Show Source):
You can put this solution on YOUR website! ~ = Square root
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(2x-7)(x+1)=-4
►Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
(2x*x+2x*1-7*x-7*1)=-4
►Simplify the FOIL expression by multiplying and combining all like terms.
(2x^(2)-5x-7)=-4
►Remove the parentheses around the expression 2x^(2)-5x-7.
2x^(2)-5x-7=-4
►Since -7 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 7 to both sides.
2x^(2)-5x=7-4
►Subtract 4 from 7 to get 3.
=► 2x^(2)-5x=3
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-12=2(x+4)^(2)
►Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
2(x+4)^(2)=-12
►Divide each term in the equation by 2.
(2(x+4)^(2))/(2)=-(12)/(2)
►Simplify the left-hand side of the equation by canceling the common terms.
(x+4)^(2)=-(12)/(2)
►Simplify the right-hand side of the equation by simplifying each term.
(x+4)^(2)=-6
►Take the square root of each side of the equation to setup the solution for x.
~((x+4)^(2))=\~(-6)
►Remove the perfect root factor (x+4) under the radical to solve for x.
(x+4)=\~(-6)
►Pull all perfect square roots out from under the radical. In this case, remove the i because it is a perfect square.
(x+4)=\i~(6)
►First, substitute in the + portion of the \ to find the first solution.
(x+4)=i~(6)
►Remove the parentheses around the expression x+4.
x+4=i~(6)
►Since 4 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4 from both sides.
x=-4+i~(6)
►Next, substitute in the - portion of the \ to find the second solution.
(x+4)=-i~(6)
►Remove the parentheses around the expression x+4.
x+4=-i~(6)
►Since 4 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4 from both sides.
x=-4-i~(6)
►The complete solution is the result of both the + and - portions of the solution.
x=-4+i~(6),-4-i~(6)
►This is the trigonometric form of a complex number where |z| is the modulus and T is the angle created on the complex plane.
z=a+bi=|z|(cosT+isinT)
►The modulus of a complex number is the distance from the origin on the complex plane.
|z|=~(a^(2)+b^(2)) where z=a+bi
►Substitute the actual values of a=-4 and b=~(6).
|z|=~(((~(6))^(2)+(-4)^(2)))
►A radical can be rewritten as a factor with a fractional exponent. The exponent is written as the inverse of the index of the radical, in this case (1)/(2).
|z|=~((((6)^((1)/(2)))^(2))+(-4)^(2))
►Expand the exponent of 2 to the inside factor (6)^((1)/(2)). To expand the exponent, multiply the exponents together.
|z|=~((((6)^(((1)/(2))(2))))+(-4)^(2))
►Remove the parentheses around the expression 6.
|z|=~(((6))+(-4)^(2))
►Remove the parentheses around the expression 6.
|z|=~((6)+(-4)^(2))
►Squaring an expression is the same as multiplying the expression by itself 2 times.
|z|=~((6)+((-4)(-4)))
►Multiply -4 by -4 to get 16.
|z|=~((6)+((16)))
►Remove the parentheses around the expression 16.
|z|=~((6)+(16))
►Remove the parentheses that are not needed from the expression.
|z|=~(6+16)
►Add 16 to 6 to get 22.
|z|=~(22)
►The angle of the point on the complex plane is the inverse tangent of the complex portion over the real portion.
T=tan^(-1)((~(6))/(-4))
►Since inverse tangent of ((~(6)))/((-4)) produces an angle in the 2nd quadrant, the value of the angle is .
T=
►Substitute the values of T= and |z|=~(22).
~(22)(cos()+isin())
►Replace the right-hand side of the equation with the trigonometric form.
x=~(22)(cos()+isin())
►Use De Moivre's Theorem to find an equation for x.
r(cos(1T)+isin(1T))=-4+i~(6),-4-i~(6)=~(22)(cos()+isin())
►Equate the modulus of the trigonometric form to r to find the value of r.
r=
►Find the possible values of T.
cos(1T)=cos(2`n) and sin(1T)=sin(2`n)
►Finding all the possible values of T leads to the equation 1T=2`n.
1T=2`n
►Find the value of T for r=0.
For n=0: 1T=2`(0)
►Solve the equation for T.
For n=0: T=0
►Use the values of T and r to find a solution to the equation x=-4+i~(6),-4-i~(6).
x0=(cos(0)+isin(0))
►Convert the solution to rectangular form.
=► x0=1
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