Solve for x for the following equation: 3(x-2)^2+2(x+1)(x-1)=x^2+4x+11

Question

Step 1:

First term:

(x-2)^2=x^2-4x+4

3(x-2)^2=3(x^2-4x+4)=3x^2-12x+12

Second term:

(x+1)(x-1)=x^2-1 (difference of squares)

2(x^2-1)=2x^2-2

Left hand side becomes:

3x^2-12x+12+2x^2-2=5x^2-12x+10

Right hand side:

x^2+4x+11

Step 2

5x^2-12x+10=x^2+4x+11

Step 3:

Move all the terms to one side

5x^2-12x+10-x^2-4x-11=0

(5x^2-x^2)+(-12x-4x)+(10-11)=0

4x^2-16x-1=0

Step 4

solve using the quadratic formula

x=-(-16)±√(-16)^2-4(4)(-1)/2(4)

x=16±√256+16/8

x=16±√272/8

√272=√16*17=4√17

x=16±4√17/8

x=4(4±√17/8)=4±√17/2

FINAL ANSWER:

x=4±√17/2

Answer 1

In place of your step 4:

Answer 2

3(x-2)^2 +2(x+1)(x-1) = x^2 +4x +11

3(x^2 -4x+4) +2(x^2-1) = ditto

3x^2 -12x+12 +2x^2 -2 = ditto

5x^2 -12x +10 = x^2 +4x +11

4x^2 -16x -1 = 0

use quadratic formula: x=-b/2a +/-(1/2a)sqr(b^2-4ac)

x=16/8 +/-(1/8)sqr(16^2 +16)

= 2 +/-(1/2)sqr17

or

complete the square

4(x^2 -4x +4) = 1 +16

(x-2)^2 = 17/4

x-2 = +/-(sqr17)/2

x = 2+/-.5sqr17

Answer 3

3(x-2)² + 2(x+1)(x-1) = x² + 4x + 11

There are a few approaches to consider, all valid. When you are pressed for time with something like this, the “brute force” method of just multiplying everything out, then regroup the “like" terms, is a good approach to avoid making mistakes when under pressure, then the Quadratic Equation at the end… which can also get messy, but OK.

With the luxury of a bit more time (like homework, and practice), it may also pay to look for patterns you already recognize… with an eye to simplification… or at least to not see it as one giant mess, but several smaller messes.

The term (x-2)² reminds us of (x - a)² = x² - 2ax + a²

(x+1)(x-1) reminds us of (x + a)(x - a) = x² - a²

Therefore, rather than multiplying everything out, which is a perfectly reasonable approach, then regrouping everything in terms of x² , x, then units… you could instead try the regrouping “upfront”, and put everything on the Left Hand Side (LHS) "as you go", as follows, with an eye to obtaining the familiar: ax² + bx + c = 0

x² terms:

We know that 3(x-2)² will produce 3x² … we don’t care about the x’s or units (for now)

We know that 2(x+1)(x-1) will produce 2x²

On the Right Hand Side (RHS), we also have x² we want to bring back to the Left Hand Side (LHS)

With all x² moved to the LHS, we get: 3x² + 2x² - x² = 4x²

x terms:

We know that 3(x-2)²… from (x - a)² = x² -2ax + a², will produce, for the ‘x’s: 3(-2ax) = 3(-4x) = -12x

We know that 2(x+1)(x-1) will produce NO ‘x’ term : same as (x - a)(x + a) = x² - a²

We also have 4x on the RHS

Regrouping all the x’s on the LHS produces: -12x - 4x = -16x

Units:

From (x - a)² = … + a², we know that 3(x-2)² will produce 3(2²) = 12

From (x - a)(x + a) = x² - a², we know that 2(x+1)(x-1) will produce 2(-1²) = -2

From the RHS, we have 11 we must move to the LHS, which becomes -11

On the LHS, we now get: 12 - 2 - 11 = -1

In summary, we have done, term by term (almost ‘mentally’), all the work to obtain: 4x² - 16x - 1 = 0 , by breaking things down into the smaller pieces.

Now:

4x² - 16x - 1 = 0 lends itself well to the Quadratic Equation… sometimes another approach can be to use the same approach that is often used to ‘prove’ the Quadratic Equation in the first place. What’s that ? By Completing the Square...

Ask yourself: Is "4x² - 16x - 1” a perfect square ? Sure doesn’t look like it, right ? I mean, (2x ± 1)(2x ± 1) will give you 4x²… and will give you “-1” for the units, but that "-16x” in the middle is not going to happen, right ?

So, we forget the "- 1", just move it (out of the way) to the RHS. We have:

4x² - 16x = 1

Now, we wonder what could we add to both sides, so that 4x² - 16x + ?? becomes a perfect square ?

Back to our (2x - a)(2x - a) = 4x² - 4ax + a² idea:

Here, our -4ax is -16x, and all we need is “a²”, i.e. “4² = 16”. Sweet… so we must add 16 to each side:

(2x - 4)² = 4x² - 16x + 16 = 1 + 16

We now have our “completed” (perfect) square:

(2x - 4)² = 17

==> 2x - 4 = ±√17 ==> 2x = 4 ± √17 ==> x = 2 ± (√17)/2

Notice that we did not use the Quadratic equation at all here… instead, we used the same technique (i.e. completing the square) that is used to ‘prove’ the Quadratic Formula in the first place. Sometimes, knowing how something is actually ‘proved’ can give you additional insight.

Well, "all roads lead to Rome”. You have many tools at your disposal in Algebra, and Truth and Logic will always get you home, even if you take some scenic detours. You do not have just one tool for these problems, but several. Get to know all your tools… each could come in handy !

Answer 4

3(x -2)² + 2(x + 1)(x - 1)= x ² + 4x + 11

3(x ² - 4x +4) + 2(x ² - 1) = x ² + 4x + 11

3x ² -12x + 12 + 2x ² -2 = x ² + 4x + 11

5x ² - 12x + 10 = x ² + 4x + 11

4x ² - 16x - 1 = 0

X ² - 4x - 1/4 = 0

Solve by completing the square.

X ² -4x = 1/4

X ² - 4x + 4 = 17/4

(x - 2)² = 17/4

√ (x -2)² = ± √ (17/4)

(x -2) = ± (√ 17)/2

X = 2 ± (√ 17)/2