8x 4 2 4x 4

$4 \exponential{ten}{2} + 8 10 + 4 = ane $

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4x^{ii}+8x+four-1=0

Subtract 1 from both sides.

4x^{2}+8x+3=0

Subtract 1 from 4 to become three.

a+b=viii ab=4\times iii=12

To solve the equation, factor the left manus side by grouping. Kickoff, left hand side needs to be rewritten as 4x^{2}+ax+bx+three. To find a and b, fix a organisation to exist solved.

1,12 ii,6 3,4

Since ab is positive, a and b have the aforementioned sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

1+12=thirteen 2+6=8 three+4=7

Calculate the sum for each pair.

a=two b=half-dozen

The solution is the pair that gives sum 8.

\left(4x^{two}+2x\right)+\left(6x+3\right)

Rewrite 4x^{2}+8x+3 as \left(4x^{two}+2x\right)+\left(6x+iii\correct).

2x\left(2x+1\correct)+three\left(2x+ane\right)

Gene out 2x in the offset and 3 in the 2nd group.

\left(2x+1\right)\left(2x+3\right)

Factor out mutual term 2x+ane by using distributive belongings.

x=-\frac{1}{2} x=-\frac{3}{2}

To discover equation solutions, solve 2x+1=0 and 2x+iii=0.

4x^{2}+8x+four=1

All equations of the form ax^{two}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{ii}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when information technology is subtraction.

4x^{2}+8x+4-i=1-ane

Subtract 1 from both sides of the equation.

4x^{2}+8x+4-i=0

Subtracting one from itself leaves 0.

4x^{two}+8x+3=0

Subtract i from 4.

x=\frac{-eight±\sqrt{eight^{2}-4\times iv\times 3}}{2\times 4}

This equation is in standard class: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and three for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-8±\sqrt{64-4\times four\times 3}}{2\times four}

Square 8.

x=\frac{-8±\sqrt{64-sixteen\times 3}}{2\times 4}

Multiply -4 times 4.

ten=\frac{-8±\sqrt{64-48}}{2\times four}

Multiply -16 times 3.

10=\frac{-eight±\sqrt{sixteen}}{2\times 4}

Add 64 to -48.

10=\frac{-viii±4}{2\times iv}

Have the square root of 16.

x=\frac{-8±4}{8}

Multiply ii times 4.

10=\frac{-four}{8}

At present solve the equation x=\frac{-8±4}{eight} when ± is plus. Add -8 to 4.

10=-\frac{1}{2}

Reduce the fraction \frac{-4}{viii} to everyman terms by extracting and canceling out iv.

x=\frac{-12}{8}

Now solve the equation x=\frac{-eight±4}{8} when ± is minus. Subtract 4 from -8.

x=-\frac{3}{2}

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out four.

x=-\frac{1}{2} x=-\frac{3}{ii}

The equation is now solved.

4x^{2}+8x+iv=ane

Quadratic equations such equally this i can exist solved by completing the square. In order to consummate the foursquare, the equation must start exist in the class x^{2}+bx=c.

4x^{2}+8x+four-iv=1-4

Subtract 4 from both sides of the equation.

4x^{ii}+8x=1-4

Subtracting 4 from itself leaves 0.

4x^{two}+8x=-3

Subtract 4 from 1.

\frac{4x^{2}+8x}{4}=\frac{-iii}{4}

Divide both sides by 4.

x^{2}+\frac{eight}{4}x=\frac{-iii}{iv}

Dividing by four undoes the multiplication by 4.

ten^{ii}+2x=\frac{-3}{4}

Divide 8 past four.

x^{2}+2x=-\frac{iii}{four}

Divide -3 by 4.

x^{two}+2x+1^{2}=-\frac{iii}{4}+ane^{2}

Divide ii, the coefficient of the x term, past two to get 1. 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+ane=-\frac{3}{4}+1

Square 1.

x^{2}+2x+i=\frac{ane}{4}

Add -\frac{3}{four} to 1.

\left(x+one\right)^{2}=\frac{1}{four}

Factor x^{2}+2x+one. In general, when x^{two}+bx+c is a perfect square, it can always exist factored as \left(x+\frac{b}{ii}\right)^{2}.

\sqrt{\left(x+1\correct)^{2}}=\sqrt{\frac{1}{4}}

Take the square root of both sides of the equation.

x+ane=\frac{1}{2} x+1=-\frac{one}{2}

Simplify.

ten=-\frac{1}{2} 10=-\frac{3}{two}

Subtract 1 from both sides of the equation.