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

Question

I know the answer is 2,-10  can you please help me understand how to get that. thank you

Answer 1

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

Answer 2

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