(x-1)(x-2)zo, XER.
9.
Solve
(x-3)(x-4)
Answers
Step-by-step explanation:
Firstly, expand the brackets to produce a quartic expression.
Then, make the expression equal zero by subtracting 3024 from both sides. You should get:
x^4+10x^3+35x^3+50x^2–3000=0
From this expression, you can see that almost all the terms’ coefficients are multiples of five. That means that, for the expression to equal zero, x^4 must also be a multiple of five.
Substitute multiples of five into this equation. You will quickly discover that the equation is solved when x=5. This means that, if we re-factorise this expression, (x-5) will be a factor.
Divide the expression through by (x-5) and you should get:
x^3+15x^2+110x +600=0
We have removed one solution, and can now move on to the others.
This expression again implies a solution that is a multiple of five, but we can also see that all of the coefficients are positive. This means that, for the equation to be solved, some terms must be negative. Therefore, substitution of negative multiples of five should bring about another solution. Minus 10 will solve this equation, and therefore, (x+10) is a factor of the expression.
Removing yet another factor, we are left with another expression:
x^2+5x+60=0
Going back, we have discovered that the initial equation can be written as:
(x-5)(x+10)(x^2+5x+60)=0
The quadratic has no real solutions, but has two complex ones, found using the quadratic equation:
(-b+-(sqrtb^2–4ac))÷(2a)=x where
ax^2+bx+c=0
Therefore the solutions are:
x=5, x=-10 and x=(-5+-(sqrt215)i)÷2
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