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# Completing squares

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What is (c+1)^2= 21/2

Guest May 19, 2017

#2
+77243
+1

(c+1)^2= 21/2     take pos/neg roots

c + 1   =  ±√(21/2)      subtract 1 from both sides

c  =   ±√(21/2)  -  1

+√(21/2)  -  1  ≈  2.24

-√(21/2)  -  1  ≈  -4.24

CPhill  May 19, 2017
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#1
+210
+1

For left part =  (21)/(2):

Because of the laws of exponents, we can take the square root of each side of the equation, and still end up with the correct answer: $$(c+1)^2=\frac{21}{2} \rightarrow c+1=\sqrt{\frac{21}{2}}$$

So $$c = \sqrt{\frac{21}{2}}+1$$

For left part = 2+(1/2):

We can still use the laws of exponents: $$(c+1)^2=\frac{5}{2} \rightarrow c+1 = \sqrt{\frac{5}{2}}$$

Simplify:

$$c=\sqrt{\frac{5}{2}}+1$$

helperid1839321  May 19, 2017
#2
+77243
+1

(c+1)^2= 21/2     take pos/neg roots

c + 1   =  ±√(21/2)      subtract 1 from both sides

c  =   ±√(21/2)  -  1

+√(21/2)  -  1  ≈  2.24

-√(21/2)  -  1  ≈  -4.24

CPhill  May 19, 2017
#3
+1224
0

$$c=\pm\frac{\sqrt{42}}{2}-1$$

$$c\approx2.24$$ or $$c\approx-4.24$$

Let's see why. Your original equation in this:

$$(c+1)^2=\frac{21}{2}$$     Take the square root of both sides

$$c+1=\pm\sqrt{\frac{21}{2}}$$     In quadratics, there are always 2 answers, hence plus or minus. First, let's place the square root of 21/2 in simplest radical form:

$$\sqrt{\frac{21}{2}}=\frac{\sqrt{21}}{\sqrt{2}}$$     First, distribute the radical to both the numerator and denominator. Notice how there is a radical in the denominator, which means we must rationalize it.

$$\frac{\sqrt21}{\sqrt2}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{42}}{2}$$     To rationalize this fraction, we multiply the fraction by the square root of 2 over the square root of two. This is the equivalent of multiplying the fraction by one, so the fraction's value is not being changed. 42 has no perfect square factors, so the radical is left as is.

$$c+1=\pm\frac{\sqrt{42}}{2}$$Subtract one on both sides to get the final answer.

$$c=\pm\frac{\sqrt{42}}{2}-1$$

TheXSquaredFactor  May 20, 2017

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