How do you solve a quadratic equation thats not straightforward?

Question

How would you factorise and solve an equation in these forms e.g (x - 2)2 = x or 6(x-1)2 = 5x - 6 or x + 4 = 36/x - 1

Really confused! Please help!

Answer 1

Take it one step at a time. Expand your brackets, collapse similar terms, rinse, repeat.

Note that when trying to write squared where you just have plain text, write, e.g x^2 (I presume (x - 2)2 means (x -2)^2, i.e. (x-2)(x-2) ). This should make things a bit clearer.

Remember your rules for handling equations, in the first two, once you've expanded, just subtract the RHS from the LHS, then simplify. For the third one, I'd change that to (x +4)(x-1)=36, expand, simplify, then subtract the 36 from both sides.

Answer 2

(x - 2)2 = x
or 2x - 2*2 = x
or 2x - x = 4
or x = 4

This is not a quadratic equation and so is the second one. But the the third one is a quadratic equation. So I am doing that for you.

I assume that the equation is
x + 4 = 36/(x -1)
and not
x + 4 = (36/x) -1
because it is not clear from your writing. The equation which you have written means the second one but that doesn't make any sense as then the equation could be given as x + 3 = 36/x.

x + 4 = 36/(x -1)
or (x + 4)(x - 1) = 36
or x^2 + 3x - 40 = 0
or x^2 + 8x - 5x - 40 = 0
or x(x + 8) - 5(x + 8) = 0
or (x - 5)(x + 8) = 0

Therefore either x = 5 or x = -8

Answer 3

you just need to transform these into the standard quadratic equation to sovle them although the first two only have a first degree x term so they really aren't quadratic equations:

(x - 2)2 = x
Distribute the 2:(2x-4)=x
Add 4 to each side, subtract x from each side:x=4

6(x-1)2 = 5x - 6
Distribute the 6 and the 2: (12x-12)=(5x-6)
Add 12 to both sides, subtract 5x from both sides: 7x=6
divide both sides by 7: x=6/7

x + 4 = 36/x - 1
multiply both sides by x/36: x^2/36 +4x/36=1 -x/36
Add x/36 and subtract 1 from both sides of the equation:
x^2/36+5x/36-1=0
multiply by 36: x^2+5x-36=0
factor:(x+9)(x-4)=0
x={-9,4}
then, since we introduced an extra x term by multiplying by x/36, we need to check that both our answers work:

if x=4:
x + 4 = 36/x - 1
4+4=36/4-1
8=9-1
8=8; therefore, 4 is an answer to the equation.

if x=-9
-9+4=36/(-9)-1
-5=-4-1
-5=-5; therefore, -9 is also an answer to the equation.
x={4,-9}

Answer 4

Simplify the equation to group like terms and then use the quadratic equation.

For example, your third equation x + 4 = 36 / (x - 1)
Multiply both sides by (x-1)/(x-1), you get:
(x - 1) * (x + 4) = 36
Multiply out the terms on the left side to get:
x^2 + 3x -4 = 36
Now collect into the quadratic equation form:
x^2 + 3X + 32 = 0
Now use the quadratic equation to solve
x = [-b +- sqrt(b^2 - 4ac)]/2a
For this example, a = 1, b = 3, and c = 32.

Answer 5

easy. you transform them in order to get something in the form ax^2 + bx + c =0, which you can solve using the formula:

x =[ -b +/- sqrt(b^2 - 4ac)]/2a

take the first one:
(x-2)^2 =x

you develop, you get:
x^2-4x+4=x

you subtract x from each side:
x^2 - 5x + 4 = 0

using the formula above you'd get:

x = [5 +/- sqrt(25-16)] / 2
= [5 +/- 3] / 2

so x1=4, and x2=1


do the same for the other two, i.e. develop the squares or multiplications, simplify, regroup terms so that it takes the form ax^2 + bx + c = 0


Good luck

Answer 6

Very simple. If the equation is not straight forward and cannot be factorized, don't factorize.
Let us say your equation is
ax^2+bx+c=0
Then it has two solutions
1............x = (-b+square root of (b^2-4ac)./2a
2............x = (-b-square root of (b^2-4ac)/2a

Example
Say
9x^2+3x-10=0
Here a=9, b=3 and c= -10
Therefore the first answer is
x= (-3+square root of (9-(4*9*(-10))))/2*9
= (-3+square root of (9+360))/18
= (-3+square root of 369)/18

Answer 7

very easy:
first simplify it,
x+4=36/x-1
x+4=(36-x)/x
x^2+4x=36-x
x^2+5x-36=0

then factorize;
x^2+9x-4x-36=0
x(x+9)-4(x+9)=0
(x-4)(x+9)=0
x=4 or -9

Answer 8

In each case, start by multiplying both sides by x.

Answer 9

Use MDAS..

Specifically Multiply, divide, addition, subtration....

ex..
multiply(divide, add and subtract) all that needs to be multiplied(divide,add and subtract)........ ano so on and so forth....