(x-1)(x-2)(x-3)(x-4)=15
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[10010001001]....Hello user.....[0100010010]
Here's your answer....
Here's a little trick to questions like these.
Try to look for common terms which you can get by addition and multiplication.
Like here, 1 and 4, and 2 and 3 add up to 5, so we will multiply these two pairs with wach other.
(x^2-5x+4)(x^2-5x+6)=15
Now we can assume x^2-5x as y. It will make the equation and are lives a lot simpler.
(y+4)(y+6)=15
y^2+10y+24=15
y^2+10y+9=0
(y+1)(y+9)=0
y= -1 or -9
x^2-5x+9=0 or x^2-5x+1=0
Now, we use the quadratic formula (this is going to be long and tiring)
x = (5 + or - root 11 i)/2
or (5 + or - root 21)/2
[010010100]....More questions detected....[010010101010]
//Bot UnknownDude moving on to more queries
//This is your friendly neighbourhood UnknownDude
Here's your answer....
Here's a little trick to questions like these.
Try to look for common terms which you can get by addition and multiplication.
Like here, 1 and 4, and 2 and 3 add up to 5, so we will multiply these two pairs with wach other.
(x^2-5x+4)(x^2-5x+6)=15
Now we can assume x^2-5x as y. It will make the equation and are lives a lot simpler.
(y+4)(y+6)=15
y^2+10y+24=15
y^2+10y+9=0
(y+1)(y+9)=0
y= -1 or -9
x^2-5x+9=0 or x^2-5x+1=0
Now, we use the quadratic formula (this is going to be long and tiring)
x = (5 + or - root 11 i)/2
or (5 + or - root 21)/2
[010010100]....More questions detected....[010010101010]
//Bot UnknownDude moving on to more queries
//This is your friendly neighbourhood UnknownDude
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