what is the answer for
Solve x2 + 8x – 48

Question

what is the answer for
Solve x2 + 8x – 48

anonymous · Answer

is anyone this

ivettef365 · Answer

you need to find a multiple of 48 than when you subtract = 8
that would be 12 x 4, can you take it from there

anonymous · Answer

how do i do that

ivettef365 · Answer

now that you have the numbers that you need you apply it like this
(x+12)(x-4)

anonymous · Answer

ok

whpalmer4 · Answer

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.