Question 3457
You have a few options for solving these problems.<br>

<ol>
<li>x^2-4=0</li>
<li>x^2+x=0</li>
<li>x^2+7x+12=0</li>
<li>x^2+5x=-4</li>
<li>3x^2+5x-10=0</li>
<li>4x^2+3x+8=0</li>
</ol>
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In the first problem, you can subtract zero from both sides to get {{{x^2 = 4}}} or x = +/- 2, or<br>
you can factor the problem {{{x^2 - 4 = (x + 2)(x - 2) = 0}}}<br>
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In the second problem, you can factor x from the equation to get<br>
{{{x(x + 1) = 0}}}, this gives us x = 0 and x = -1<br>
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In this problem, you have to make some observations, first notice you can factor 12 into the following pairs ({12, 1}, (6, 2} and {4, 3}). The pair {4, 3} sum to seven, so we can factor this equation as follows:<br>
{{{x^2 + 7x + 12 = (x + 4)(x + 7) = 0}}}, therefore x = -4 and x = -7.<br>
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This problem is similar to the previous problem, but first you must subtract 4 from both sides to get x^2 + 5x + 4 = 0. The common factors of 4 are ({4,1} and {2,2}). The pair {4,1} sum to 5, so we can factor this equation as follows:<br>
{{{x^2 + 5x + 4 = (x + 4)(x + 1) = 0}}}, therefore x = -1 and x = -4.<br>
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This problem is a bit more complex, it requires the use of the quadratic formula for finding roots of a quadratic equation with the form {{{(ax^2 + bx + c = 0}}}. If the roots are x<sub>1</sub> and x<sub>2</sub>, the quadratic formula says:<br>
x<sub>1</sub> = {{{(-b + sqrt( b^2-4*a*c ))/(2*a) }}} and x<sub>2</sub> = {{{(-b - sqrt( b^2-4*a*c ))/(2*a) }}}<br>
For 3x^2 + 5x - 10 = 0, a = 3, b = 5 and c = -10. This means the roots are:<br>
x<sub>1</sub> = {{{(-5 + sqrt( 5^2-4*3*(-10) ))/(2*3) }}}, which reduces to, x<sub>1</sub> = {{{(-5 + sqrt( 25+120 ))/6 = (-5 + sqrt( 145 ))/6 }}}
and x<sub>2</sub> = {{{(-5 - sqrt( 5^2-4*3*(-10) ))/(2*3) }}}, which reduces to, x<sub>2</sub> = {{{(-5 - sqrt( 25+120 ))/6 = (-5 - sqrt( 145 ))/6 }}}
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Like the previous problem, this one requires use of the quadratic equation, unfortunately, this one involves complex numbers<br>
The equation 4x^2 + 3x + 8 = 0 gives us a = 4, b = 3 and c = 8<br>
The roots of this equation are:<br>
x<sub>1</sub> = {{{(-4 + sqrt( 4^2-4*4*8 ))/(2*4) }}}, which reduces to x<sub>1</sub> = {{{(-4 + sqrt( 4^2-4*4*8 ))/(2*4) = (-4 + sqrt( -112 ))/8 = (-4 + sqrt(112)i)/8}}}<br>
and x<sub>2</sub> = {{{(-4 - sqrt( 4^2-4*4*8 ))/(2*4) }}}, which reduces to x<sub>2</sub> = {{{(-4 - sqrt( 4^2-4*4*8 ))/(2*4) = (-4 - sqrt( -112 ))/8 = (-4 - sqrt(112)i)/8}}}<br>