How do you solve # 2x^2 - 7x - 9 = 0#?

2 Answers
Gió
Jul 2, 2015

I found:
#x_1=-1#
#x_2=9/2#

Explanation:

We can use the Quadratic Formula but...it is boring!
Try this instead:
Divide by #2#:
#x^2-7/2x-9/2=0#
Rearrange:
#x^2-7/2x=9/2#
Add ans subtract #49/16#;
#x^2-7/2x+49/16-49/16=9/2#
Rearrange:
#x^2-7/2x+49/16=9/2+49/16#
#(x-7/4)^2=(72+49)/16=121/16#
#x-7/4=+-sqrt(121/16)=+-11/4#
So:
#x_1=7/4-11/4=-1#
#x_2=7/4+11/4=-18/4=9/2#

Nghi N.
Jul 2, 2015

Solve #y = 2x^2 - 7x - 9 = 0#

Explanation:

To solve this type of quadratic equation, use shortcut. It can save you a lot of time and work.
(a - b + c = 0) --> 2 real roots: x = - 1, and #x = -c/a = 9/2#

Remind of Shortcut:

  1. When (a + b + c = 0) -> 2 real roots: 1 and #(c/a)#

  2. When (a - b + c = 0 -> 2 real roots: - 1 and #(-c/a)#

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