3 ways to find the solutions of \(x^2+8x-48=0\):
1) factor
2) complete the square
3) quadratic formula
To factor a quadratic of the form \(x^2+bx+c=0\), you need to find a pair of numbers that when multiplied give you \(c\) and when added give you \(b\). If you multiply two binomials together:
\[(x+a)(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab\]you can see that the constant term, \(ab\), is formed by the product of the two constants in the binomials. Furthermore, the coefficient of the middle term, \((a+b)x\), comes from the sum of the two constants in the binomials. The situation is a bit more complicated if the \(x\) terms have a coefficient other than 1, but we'll leave that for another day.
So, we need to find a pair of numbers \(a, b\) such that \(a+b = 8\) and \(a*b = -48\). If we look at the factors of -48, we'll eventually come across -4 and 12, which satisfy the constraints. \[-4+12 = 8\]\[ -4*12 = -48\]and \[(x-4)(x+12) = x^2+12x-4x-48 = x^2+8x-48\]
Great, we've factored it, how does that help? Well, the solutions that make the quadratic equal 0 are the solutions to \(x-4 = 0\) and \(x+12 = 0\). Solve the two trivial equations and you've got your answers.
But what if you can't find some numbers that work? That brings us to:
2) Completing the square
If we can rewrite the left hand side of the equation to be a perfect square, then we can take the square root of both sides of the equation and find our two solutions. Here's how we do it:
\[x^2+8x-48=0\]rewrite so only the terms containing \(x\) are on the left side\[x^2+8x=48\]Now we take half of the coefficient of the \(x\) term, square it, and add that quantity to both sides of the equation. As long as we add it to both sides, we preserve the equality.
\[x^2+8x+(\frac{8}{2})^2 = 48+(\frac{8}{2})^2\]which simplifies to \[x^2+8x+16=64\]Now we can rewrite the left hand side as a perfect square, remembering that \[(x+a)(x+a) = x^2+ax +ax + a^2 = x^2 + 2ax + a^2\]which means the left hand side can be written as \((x+4)(x+4) = (x+4)^2\) giving us
\[(x+4)^2 =64\]Take the square root of both sides to get\[(x+4) = \pm 8\]Now we solve \[x+4=8\] and \[x+4=-8\]to get our two solutions.
That's a bunch of algebra, and you might wonder if there's a way you can solve it once and for all. There is:
3) Quadratic formula
If you have a quadratic of the form \(ax^2+bx+c=0, a\ne0\) then you can simply plug the coefficients \(a,b,c\) into this formula to get the solutions:
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Here we would have \(a = 1, b = 8, c = -48\) and our solutions would be
\[x=\frac{-8\pm\sqrt{8^2-4(1)(-48)}}{2(1)} = \frac{-8\pm\sqrt{64+192}}{2} = \frac{-8\pm16}{2} = -4\pm8\]
If you want a learning exercise, take the polynomial \(ax^2+bx+c=0\) and complete the square on it. You'll get the quadratic formula as a result.