Question

which of the following is equivalent to the equation below by completing the square?
$x^2 + 4x - 72 = 0$
$square\\ (x + 4)^2 = 88$
$square\\ (x + 4)^2 = 72$
$square\\ (x + 2)^2 = -72$
$square\\ (x + 2)^2 = 76$
what are the solutions to $x^2 + 4x - 72 = 0$?

Explanation:

Step1: Start with the equation

We have the quadratic equation \(x^{2}+4x - 72=0\). First, we isolate the \(x\)-terms: \(x^{2}+4x=72\).

Step2: Complete the square

To complete the square for the expression \(x^{2}+4x\), we take half of the coefficient of \(x\) (which is \(4\)), so \(\frac{4}{2} = 2\), and then square it: \(2^{2}=4\). We add this to both sides of the equation: \(x^{2}+4x + 4=72 + 4\).

Step3: Rewrite as a perfect square

The left - hand side can be written as a perfect square: \((x + 2)^{2}=76\) (this is the step of completing the square to get the equivalent equation). Now, to solve for \(x\), we take the square root of both sides: \(x+2=\pm\sqrt{76}\). Simplify \(\sqrt{76}=\sqrt{4\times19} = 2\sqrt{19}\approx\pm8.72\). Then \(x=-2\pm\sqrt{76}\). We can also simplify \(\sqrt{76} = 2\sqrt{19}\), or we can solve it by factoring (after completing the square or directly). Let's go back to the equation \((x + 2)^{2}=76\), so \(x+2=\sqrt{76}\) or \(x + 2=-\sqrt{76}\). \(x=-2+\sqrt{76}\approx-2 + 8.72=6.72\) and \(x=-2-\sqrt{76}\approx-2-8.72=-10.72\). Alternatively, from the factored form (if we factor the original quadratic: \(x^{2}+4x - 72=(x + 12)(x - 8)=0\), so \(x=-12\) or \(x = 8\). Wait, there is a mistake in the previous calculation of \(\sqrt{76}\). Wait, \(76 = 4\times19\), but if we use the completed square \((x + 2)^{2}=76\), then \(x=-2\pm\sqrt{76}\), but \(\sqrt{76}\approx8.7178\), but let's check with factoring. Let's factor \(x^{2}+4x - 72\). We need two numbers that multiply to \(-72\) and add up to \(4\). The numbers are \(12\) and \(-8\) (since \(12\times(-8)=-72\) and \(12+( - 8)=4\)). So \(x^{2}+4x - 72=(x + 12)(x - 8)=0\). Setting each factor equal to zero: \(x+12 = 0\) gives \(x=-12\) and \(x - 8=0\) gives \(x = 8\). Let's check with the completed square: \((x + 2)^{2}=76\), if \(x = 8\), then \((8 + 2)^{2}=100 eq76\). Wait, I made a mistake in the completing the square step. Wait, original equation: \(x^{2}+4x-72 = 0\), \(x^{2}+4x=72\), add \(4\) to both sides: \(x^{2}+4x + 4=72 + 4=76\), so \((x + 2)^{2}=76\) is correct. But when we factor \(x^{2}+4x-72\), we have \((x + 12)(x - 8)=x^{2}-8x+12x - 96=x^{2}+4x-96 eq x^{2}+4x - 72\). Oh! I made a factoring mistake. Let's find two numbers that multiply to \(-72\) and add to \(4\). Let the numbers be \(a\) and \(b\), \(a\times b=-72\), \(a + b = 4\). Solving the system: \(b = 4 - a\), so \(a(4 - a)=-72\), \(4a-a^{2}=-72\), \(a^{2}-4a - 72=0\). Using quadratic formula for \(a\): \(a=\frac{4\pm\sqrt{16+288}}{2}=\frac{4\pm\sqrt{304}}{2}=\frac{4\pm4\sqrt{19}}{2}=2\pm2\sqrt{19}\). Wait, that's not helpful. Let's go back to the completed square. \((x + 2)^{2}=76\), so \(x=-2\pm\sqrt{76}\), \(\sqrt{76}=\sqrt{4\times19}=2\sqrt{19}\approx8.7178\). So \(x=-2 + 2\sqrt{19}\approx-2+8.7178 = 6.7178\) and \(x=-2-2\sqrt{19}\approx-2 - 8.7178=-10.7178\). Wait, but let's check by plugging \(x = 6\) into the original equation: \(6^{2}+4\times6-72=36 + 24-72=-12 eq0\). \(x = 8\): \(8^{2}+4\times8-72=64 + 32-72=24 eq0\). \(x=-12\): \((-12)^{2}+4\times(-12)-72=144-48 - 72=24 eq0\). I see my factoring mistake. Let's do factoring correctly. \(x^{2}+4x-72\), we need two numbers \(m\) and \(n\) such that \(m\times n=-72\) and \(m + n = 4\). Let's list the factor pairs of \(-72\): \((-1,72)\) sum \(71\); \((-2,36)\) sum \(34\); \((-3,24)\) sum \(21\); \((-4,18)\) sum \(14\); \((-6,12)\) sum \(6\); \((-8,9)\) sum \(1\); \((8,-9)\) sum \(-1\); \((12,-6)\) sum \(6\); \((18,-4)\) sum \(14\); \((24,-3)\) sum \(21\); \((36,-2)\) sum \(34\); \((72,-1)\) sum \(71\). Wait, there is no factor pair that adds up to \(4\)? That means the quadratic doesn't factor nicely, so we must use completing the square or quadratic formula. The quadratic formula for \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(x^{2}+4x-72=0\), \(a = 1\), \(b = 4\), \(c=-72\). Then \(x=\frac{-4\pm\sqrt{16+288}}{2}=\frac{-4\pm\sqrt{304}}{2}=\frac{-4\pm4\sqrt{19}}{2}=-2\pm2\sqrt{19}\approx-2\pm8.7178\), so \(x\approx6.7178\) or \(x\approx-10.7178\). And from the completed square \((x + 2)^{2}=76\), taking square roots: \(x+2=\pm\sqrt{76}\), so \(x=-2\pm\sqrt{76}\), which is the same as the quadratic formula result.

Answer:

To solve \(x^{2}+4x - 72=0\):

Step 1: Complete the square

1. Isolate the \(x\) - terms: \(x^{2}+4x=72\)
2. Complete the square for \(x^{2}+4x\):

• Take half of the coefficient of \(x\) (i.e., \(\frac{4}{2}=2\)) and square it (\(2^{2} = 4\)).
• Add this value to both sides of the equation: \(x^{2}+4x + 4=72 + 4\)
• Rewrite the left - hand side as a perfect square: \((x + 2)^{2}=76\)

Step 2: Solve for \(x\)

1. Take the square root of both sides: \(x + 2=\pm\sqrt{76}\)
2. Solve for \(x\):

• \(x=-2+\sqrt{76}\) or \(x=-2-\sqrt{76}\)
• Simplify \(\sqrt{76}=\sqrt{4\times19}=2\sqrt{19}\), so \(x=-2\pm2\sqrt{19}\approx-2\pm8.72\)
• \(x\approx-2 + 8.72 = 6.72\) and \(x\approx-2-8.72=-10.72\) (approximate values) or the exact form \(x=-2\pm2\sqrt{19}\)

The solutions are \(x=-2 + 2\sqrt{19}\) and \(x=-2-2\sqrt{19}\) (or approximately \(x\approx6.72\) and \(x\approx - 10.72\)).