How do you solve 2x ^ { 2} + 57x = 4972x2+57x=497?

2 Answers
Apr 12, 2017

x = 7x=7 , x=-35.5x=35.5

Explanation:

First we must convert our initial equation so that it fits satisfies ax^2 + bx + c = 0 ax2+bx+c=0.

2x^2 + 57x = 497 2x2+57x=497 becomes 2x^2 + 57x - 497 = 0 2x2+57x497=0.

From here we conclude the following:

a=2a=2
b=57b=57
c=-497c=497

Now we use the quadratic formula to find the value of xx.

x = (-b +- sqrt(b^2 -4ac))/(2a)x=b±b24ac2a

x = (-57 +- sqrt(57^2 -4(2)(-497)))/(2(2))x=57±5724(2)(497)2(2)

x = (-57 +- sqrt(3249 -8(-497)))/(4)x=57±32498(497)4

x = (-57 +- sqrt(3249 + 3976))/(4)x=57±3249+39764

x = (-57 +- sqrt(7225))/(4)x=57±72254

x = (-57 +- 85)/(4)x=57±854

x = (-57 + 85)/4 x=57+854 , x=(-57-85)/4x=57854

x = (28)/(4) x=284 , x = (-142)/(4) x=1424

x = 7x=7 , x=-35.5x=35.5

In this case it is not specified whether we want the positive x x value, so we are left with two solutions which are x = 7x=7 and x=-35.5x=35.5.

Apr 12, 2017

Given: 2x^2+57X=4972x2+57X=497

We can quickly turn this into an easily recognizable quadratic equation by subtracting 497 from both sides to get:

2x^2+57X-497=02x2+57X497=0

There are a couple of "give aways" we can see on this equation that will take us to the solution:

1) The xx exponent is ^2^2 so there are two answers for xx and there will need to be two brackets in the solution.

2) the multiplier 497497 looks like it should be divided by 77.

497/7=714977=71

Setting up the brackets: (2x ...71)(x ...7)=0

There is a + and - in the given equation so we know there will be a + and - in the brackets. We also need the larger value to be +, so that will be the 71x:

(2x + 71)(x - 7)=0 and we see that 71x - 7xx2x = 57x

Then: (2x + 71)=0

2x = - 71

x = - 71/2

And: (x-7) = 0

x = 7

To check, put the answers back into the given equation:

2x^2+57x=497

2(7)^2+57(7)=497

2(49)+399=497

98+399=497

And the other factor:

2x^2+57x=497

2(-71/2)^2+57(-71/2)=497

(5041/2) + (-4047/2) = 497

5041 - 4047 = 2xx497

994=994