/2xsquared-x-1/ =3?

Question

the lines are for absolute value. this is for college algebra. please show work.

Answer 1

well dear ;
f(x) = | 2x^2 -x - 1| = 3
{ explanation; if
f(x) = |x| ->
+x is for x > 0
- x is for x < 0
}

Step One;
+x is for x > 0
f(x) = + ( 2x^2 -x - 1 ) = 3
+ ( 2x^2 -x - 1 ) = 3
2x^2 -x - 1 - 3 =0
2x^2 - x - 4 { here we need to find the roots, x1 & x2 }
∆ = b^2 - 4 a c & ∆ ≥ 0 { if a = +2 , b= -1 & c = -4 }
∆ = (-1^2) - 4(2)(-4) = 1 + 32 = 33
x1 =( -b + √∆ )/ 2a = -(-1) + √33 / 4 = 1.68614
x2 =( -b - √∆ )/ 2a = -(-1) - √33 / 4 = -1.18614

Step Two;
if - x is for x < 0
- ( 2x^2 -x - 1 ) = 3
-2x^2 + x +1 - 3= 0
-2x^2 + x -2 = 0
{ here you should find x1 & x2 like i did before}
∆ = b^2 - 4 a c { if a = -2 , b= =1 & c = -2 }
∆ = 1^2) - 4 ( -2)(-2) = 1 - 16 = -15
this functions has no real result. b coz '∆ 'mubt be grater or equall Zero . ∆ ≥ 0

Good Luck.

*****{ Hey watch out the other answers, i could do what u did (Real Solutions
x = (1/4)(1 + sqrt(33)) or (1/4)(1 - sqrt(33)) ) !!!!!!
But Actulay we cant squar root form NEGATIVE NUMBERS )

Answer 2

This represents a simple quadratic equation!
2x^2-x-1=3
2x^2 - x - 4 =0
This is in the form ax^2 + bx + c = 0
By formula,
x = [-b +/-( b^2 - 4ac)]/2a
x = [1 +/-(1+32)/4
x = [1+33]/4 or x = [1-33]/4
x = 17/2 or -8
Since this is a quadratic equation the root x has two values.

Answer 3

|2x^2 - x - 1| = 3

same as saying

2x^2 - x - 1 = 3 or 2x^2 - x - 1 = -3
2x^2 - x - 4 = 0 or 2x^2 - x + 2 = 0

2x^2 - x - 4 = 0
x = (-b ± sqrt(b^2 - 4ac))/(2a)
x = (-(-1) ± sqrt((-1)^2 - 4(2)(-4)))/(2(2))
x = (1 ± sqrt(1 + 32))/4
x = (1 ± sqrt(33))/4
x = (1/4)(1 ± sqrt(33))

2x^2 - x + 2 = 0
x = (-b ± sqrt(b^2 - 4ac))/(2a)
x = (-(-1) ± sqrt((-1)^2 - 4(2)(2)))/(2(2))
x = (1 ± sqrt(1 - 16))/4
x = (1 ± sqrt(-15))/4
x = (1/4)(1 ± isqrt(15))

Real Solutions
x = (1/4)(1 + sqrt(33)) or (1/4)(1 - sqrt(33))

Complex Solutions
x = (1/4)(1 + isqrt(15)) or (1/4)(1 - isqrt(15))

Answer 4

2*x^2 -X -1=3, or 2*x^2 -X-4 = 0

x = -b(Plus or Minus)* ( Square rout(b^2-4ac)/ 2a

a=2, b= -1, c= -4

Substituting

x = (-(-1)(Plus or minus)*((Square rout(-1^2)-(4*2*(-4)))/2*2

= (1 +squaree rout (33))/4 or (+1 -squar rout (33)))/4

Substitute to get the answer.

Answer 5

Square on both sides to remove the modulus and solve the quartic equation for x.