4x 2 8x 1 0
$iv \exponential{x}{two} + 8 x + 4 = 1 $
x = -\frac{3}{2} = -1\frac{1}{2} = -1.5
x=-\frac{ane}{2}=-0.5
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4x^{ii}+8x+four-ane=0
Decrease 1 from both sides.
4x^{two}+8x+3=0
Subtract 1 from iv to get 3.
a+b=eight ab=4\times 3=12
To solve the equation, cistron the left mitt side by grouping. First, left paw side needs to be rewritten as 4x^{ii}+ax+bx+3. To notice a and b, fix a organisation to be solved.
1,12 2,6 3,4
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.
1+12=13 two+6=eight 3+4=7
Calculate the sum for each pair.
a=2 b=half dozen
The solution is the pair that gives sum 8.
\left(4x^{2}+2x\right)+\left(6x+iii\right)
Rewrite 4x^{ii}+8x+3 equally \left(4x^{two}+2x\right)+\left(6x+three\right).
2x\left(2x+1\correct)+3\left(2x+one\correct)
Gene out 2x in the kickoff and three in the 2d group.
\left(2x+1\right)\left(2x+iii\right)
Factor out common term 2x+1 by using distributive property.
x=-\frac{1}{2} x=-\frac{iii}{2}
To detect equation solutions, solve 2x+1=0 and 2x+three=0.
4x^{2}+8x+iv=1
All equations of the course ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives ii solutions, one when ± is addition and one when information technology is subtraction.
4x^{2}+8x+4-1=1-1
Subtract i from both sides of the equation.
4x^{2}+8x+4-i=0
Subtracting i from itself leaves 0.
4x^{2}+8x+iii=0
Subtract 1 from 4.
10=\frac{-eight±\sqrt{8^{two}-4\times 4\times 3}}{2\times 4}
This equation is in standard form: ax^{ii}+bx+c=0. Substitute 4 for a, 8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{ii}-4ac}}{2a}.
ten=\frac{-8±\sqrt{64-4\times 4\times iii}}{ii\times 4}
Foursquare 8.
x=\frac{-8±\sqrt{64-16\times 3}}{2\times 4}
Multiply -4 times iv.
x=\frac{-8±\sqrt{64-48}}{2\times four}
Multiply -16 times 3.
ten=\frac{-viii±\sqrt{16}}{two\times 4}
Add 64 to -48.
x=\frac{-8±four}{two\times four}
Accept the square root of xvi.
ten=\frac{-viii±4}{eight}
Multiply two times four.
10=\frac{-4}{8}
Now solve the equation ten=\frac{-8±4}{viii} when ± is plus. Add -viii to 4.
x=-\frac{one}{2}
Reduce the fraction \frac{-4}{8} to everyman terms by extracting and canceling out iv.
x=\frac{-12}{8}
At present solve the equation x=\frac{-8±4}{eight} when ± is minus. Subtract iv from -8.
x=-\frac{iii}{two}
Reduce the fraction \frac{-12}{8} to everyman terms by extracting and canceling out iv.
x=-\frac{1}{ii} x=-\frac{3}{2}
The equation is now solved.
4x^{2}+8x+4=one
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must start exist in the form x^{2}+bx=c.
4x^{2}+8x+four-4=1-4
Decrease 4 from both sides of the equation.
4x^{2}+8x=i-4
Subtracting 4 from itself leaves 0.
4x^{2}+8x=-3
Decrease four from one.
\frac{4x^{2}+8x}{4}=\frac{-3}{4}
Divide both sides by 4.
x^{2}+\frac{8}{iv}x=\frac{-3}{iv}
Dividing past iv undoes the multiplication past four.
x^{2}+2x=\frac{-3}{4}
Divide 8 past 4.
x^{ii}+2x=-\frac{three}{four}
Dissever -3 by iv.
ten^{two}+2x+1^{2}=-\frac{three}{4}+i^{two}
Divide 2, the coefficient of the x term, by 2 to get one. So add the foursquare of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=-\frac{3}{4}+1
Square ane.
ten^{2}+2x+1=\frac{1}{4}
Add together -\frac{3}{four} to ane.
\left(10+1\right)^{two}=\frac{i}{four}
Factor 10^{ii}+2x+one. In full general, when x^{2}+bx+c is a perfect square, it tin always be factored equally \left(x+\frac{b}{two}\right)^{2}.
\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{one}{4}}
Take the square root of both sides of the equation.
ten+1=\frac{one}{ii} 10+1=-\frac{one}{2}
Simplify.
10=-\frac{ane}{2} 10=-\frac{3}{ii}
Subtract 1 from both sides of the equation.
4x 2 8x 1 0,
Source: https://mathsolver.microsoft.com/en/solve-problem/4%20x%20%5E%20%7B%202%20%7D%20%2B%208%20x%20%2B%204%20%3D%201
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