Please using one of the quadratic formulas solve these: x^2-9x=0, 5x^2+14x=x+6, 4x^2-24=20x

Question



Answer 1

The quadratic formula is:

x = (-b ± (b²-4ac)^1/2)/2a, where a is the coefficient of the leading term (the term with x²), b is the coefficient for the term containing “x,” and c is the constant.

Thus,

in the first example, a = 1, b = -9, c = 0, and we get

x = (9 ± (81 - 4*1*0)^1/2)/(2*1) = (9 ± 9)/2

the two solutions acquired are x = 9, x = 0.

The second example given is somewhat more tricky because there are two terms which have coefficients for “x.” But no worry, simply add these terms together and move them to one side. In my case, I will choose to subtract “x + 6” from both sides of the equation and combine like terms. This results in the following equation:

5x² + 13x - 6 = 0

here, a = 5, b = 13, c = -6.

Plug into the quadratic formula

x = (-13 ± (13² - 4*5*(-6))^1/2)/(2*5) = (-13 ± 17)/10

x = 0.4, -3

Using a similar method as above by moving all terms to one side, the third example becomes

4x² - 20x -24 = 0

In this example, you could carry on and use the quadratic formula directly. However, we see that all of the coefficients are divisible by 4. So, we divide everything by 4 to give

x² - 5x - 6 = 0

This is a much easier problem to solve! In this case, a = 1, b = -5, and c = -6. By plugging these into the quadratic equation, we find that x = 6 and -1.