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x^2+8x-20=0 Solve using any method

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2 Answers

Hi! Since the coefficient of the x^2 is 1 (1x^2), you have a quick method to do this.
First identify the factors of 20 in pairs (20*1, 10*2, 5*4)
 
Now which pair when added or subtracted from each other gives you 8?
20+1=21, 20-1=19, 10+2=12, 10-2=8, 5+4=9, 5-4=1
 
You need the 10 and 2.
 
One must be negative so that when multiplied, you get a -20 as a result. However, when added or subtracted, you need to get a positive 8. Therefore the 2 is negative (10+ (-2) = 8)
 
So your answer is (x-2)(x+10)
 
Hope this helps
 
Mike
x^2+8x-20=0 Solve using any method
 
Easiest Method - Factoring:
 
To factor, you need two numbers that multiply together to give -20 and add together to give 8.
 
Multiples of -20 are ±20*±1, ±10*±2, ±5*±4.  Let's try +10 and -2:
 
10 + (-2) = 8
10*(-2) = -20
 
10 and -2 work, so:
 
x2+8x-20 = (x+10)(x-2) = 0
 
so x = -10 and 2
 
Quadratic Equation:
x = -(b/2a) ± [b2-4ac]1/2/2a
 
Your equation is (1)x2 + (8)x + (-20) = 0
 
x = -(8/2) ± [82-4(1)(-20)]1/2/2
x = -4 ± 12/2
x = -4 ± 6
 
x = 2, -10