6x^2-10x-16=0?

3 Answers
Jun 15, 2018

Solve the quadratic to get x=-1, 8/3x=1,83

Explanation:

I assume the question wants the solutions for xx.

Divide through to make the first coefficient 1:
x^2-5/3x-8/3=0x253x83=0

Complete the square:
(x-5/6)^2-25/36-8/3=0(x56)2253683=0
(x-5/6)^2=121/36(x56)2=12136

x=5/6+-sqrt(121/36)=5/6+-11/6x=56±12136=56±116

So x=-1, 8/3x=1,83.

Jun 16, 2018

y = 6x^2 - 10x - 16 = 0y=6x210x16=0
Since a - b + c = 0, use shortcut.
The 2 real roots are:
x = -1 ,and x = -c/a = 16/6 = 8/3x=ca=166=83

Reminder of shortcut .
If a + b + c = 0, the 2 real roots are: x = 1, and x = c/ax=ca
Example 1. Solve: 3x^2 + 7x - 10 = 03x2+7x10=0
x = 1, and x = c/a = -10/3x=ca=103
If a - b + c = 0, the 2 real roots are: x = -1, and x = -c/ax=ca
Example 2. Solve: 4x^2 - 5x - 9 = 04x25x9=0
x = -1, and x = -c/a = 9/4x=ca=94

Jun 16, 2018

Another way

Explanation:

if ax^2+bx+c=0 ax2+bx+c=0

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

In this case a = 6,b=-10, c = -16 a=6,b=10,c=16

x = ( -(-10) pm sqrt( (-10)^2 - ( 4 * 6 * (-16 ) ) ) )/(2*6) x=(10)±(10)2(46(16))26

x = (10pmsqrt(484) )/12 x=10±48412

x = (10 pm22 )/12 x=10±2212

x = {8/3 , -1 } x={83,1}