# Another square technique...

Use the technique of completing the square to transform the quadratic equation into the form (x + c)2 = a.

2x2 + 32x + 12 = 0

Thanks alot guys.

#### harish21

Use the technique of completing the square to transform the quadratic equation into the form (x + c)2 = a.

2x2 + 32x + 12 = 0

Thanks alot guys.
$$\displaystyle 2x^{2} + 32x + 12 = 0$$

$$\displaystyle 2(x^{2} + 16x + 6 = 0)$$

$$\displaystyle x^{2} + 16x + 6 = 0$$

$$\displaystyle x^2 + 2\times 8\times x + 6 = 0$$

Can you take it further? Please try

$$\displaystyle 2x^{2} + 32x + 12 = 0$$

$$\displaystyle 2(x^{2} + 16x + 6 = 0)$$

$$\displaystyle x^{2} + 16x + 6 = 0$$

$$\displaystyle x^2 + 2\times 8\times x + 6 = 0$$

Can you take it further? Please try
Would it be (x + 6)2 = 58 ?

#### harish21

Would it be (x + 6)2 = 58 ?
No.

You are left with $$\displaystyle x^2+ 2\times x \times 8 + 6 = 0$$.....(I)

So $$\displaystyle (x+c)^2 = x^2+ 2.x.c +c^2$$...(II)

if you compare I and II, you should see that c = 8.

So you need to have $$\displaystyle (x+8)^2$$, which is equal to $$\displaystyle x^2 + 2.x.8+ 64$$

now if you look at equation (I) again, you already have $$\displaystyle x^2+ 2\times x \times 8 + 6 = 0$$.

To make this $$\displaystyle x^2+ 2\times x \times 8 + 64 = 0$$,

you need to add 58 right?

so,

$$\displaystyle x^2+ 2\times x \times 8 + 6 +58 -58= 0$$

Now put the above equation in the form of $$\displaystyle (x+c)^2=a$$

No.

You are left with $$\displaystyle x^2+ 2\times x \times 8 + 6 = 0$$.....(I)

So $$\displaystyle (x+c)^2 = x^2+ 2.x.c +c^2$$...(II)

if you compare I and II, you should see that c = 8.

So you need to have $$\displaystyle (x+8)^2$$, which is equal to $$\displaystyle x^2 + 2.x.8+ 64$$

now if you look at equation (I) again, you already have $$\displaystyle x^2+ 2\times x \times 8 + 6 = 0$$.

To make this $$\displaystyle x^2+ 2\times x \times 8 + 64 = 0$$,

you need to add 58 right?

so,

$$\displaystyle x^2+ 2\times x \times 8 + 6 +58 -58= 0$$

Now put the above equation in the form of $$\displaystyle (x+c)^2=a$$
$$\displaystyle (x + 8)2 = 58$$ would be it than right?

#### harish21

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