How do you solve 2x^2 - 7x - 9 = 02x2−7x−9=0?

2 Answers
Gió
Jul 2, 2015

I found:
x_1=-1x1=−1
x_2=9/2x2=92

Explanation:

We can use the Quadratic Formula but...it is boring!
Try this instead:
Divide by 22:
x^2-7/2x-9/2=0x2−72x−92=0
Rearrange:
x^2-7/2x=9/2x2−72x=92
Add ans subtract 49/164916;
x^2-7/2x+49/16-49/16=9/2x2−72x+4916−4916=92
Rearrange:
x^2-7/2x+49/16=9/2+49/16x2−72x+4916=92+4916
(x-7/4)^2=(72+49)/16=121/16(x−74)2=72+4916=12116
x-7/4=+-sqrt(121/16)=+-11/4x−74=±√12116=±114
So:
x_1=7/4-11/4=-1x1=74−114=−1
x_2=7/4+11/4=-18/4=9/2x2=74+114=−184=92

Nghi N.
Jul 2, 2015

Solve y = 2x^2 - 7x - 9 = 0y=2x2−7x−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 = 9/2x=−ca=92

Remind of Shortcut:

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

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

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