Use the quadratic formula to solve 5x^2+2=x. Give exact solutions.?
回答 (8)
2008-11-10 9:25 am
✔ 最佳答案
ax² + bx + c = 0x = (-b±√(b²-4ac))/2a
5x² + 2 = x
5x² - x + 2 = 0
a = 5
b = -1
c = 2
x = (1±√(1-40))/10
= (1±√(-39))/10
= (1±i√39)/10
2016-05-27 3:46 pm
Step 1: Memorize the Baskara's formula: (-b +/- sqrt delta) over 2a 3x²-5x+4=0 Delta = -5² - 4.3.4 Delta = 25 -48 delta = -23 x =[ -(-5) +/- √-23] : 2*3 x' = [5 +/- √-23] : 6 x' = 5/6 + √-23/6 x" = 5/6 - √-23/6 Attention: -√23 is not equal √-23. ><
2008-11-10 9:59 am
5x^2 + 2 = x
5x^2 - x + 2 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 5
b = -1
c = 2
x = [1 ±√(1 - 40)]/10
x = [1 ±√-39]/10 (imaginary number)
(no real roots)
5x^2 - x + 2 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 5
b = -1
c = 2
x = [1 ±√(1 - 40)]/10
x = [1 ±√-39]/10 (imaginary number)
(no real roots)
2008-11-10 9:45 am
5 x ² - x + 2 = 0
x = [ 1 ± √ (1 - 40) ] / 10
x = [ 1 ± √ (- 39) ] / 10
x = [ 1 ± i √ (39) ] / 10
x = [ 1 ± √ (1 - 40) ] / 10
x = [ 1 ± √ (- 39) ] / 10
x = [ 1 ± i √ (39) ] / 10
2008-11-10 9:34 am
ax² + bx + c = 0
x = (-b±√(b²-4ac))/2a
5x² + 2 = x
5x² - x + 2 = 0
a = 5, b = -1 , c = 2
x = (1±√(1-40))/10
= (1±√(-39))/10
= (1±i√39)/10
x = (-b±√(b²-4ac))/2a
5x² + 2 = x
5x² - x + 2 = 0
a = 5, b = -1 , c = 2
x = (1±√(1-40))/10
= (1±√(-39))/10
= (1±i√39)/10
2008-11-10 9:28 am
5x² - x + 2 = 0
x = [ 1 ± √(1 - 40) ] / 10 ....................... [-b ± √(b² - 4ac)] / (2a)
= [ 1 ± i√39 ] / 10
Answer: ( 1 ± i√39 ) / 10
x = [ 1 ± √(1 - 40) ] / 10 ....................... [-b ± √(b² - 4ac)] / (2a)
= [ 1 ± i√39 ] / 10
Answer: ( 1 ± i√39 ) / 10
2008-11-10 9:27 am
5x^2 + 2 = x
5x^2 - x + 2 = 0
now the equation is in the form of ax^2 + bx + c = 0
you can use the quadratic formula
x= (-b + sqr(b^2 - 4ac)) / 2a and
x= (-b - sqr(b^2 - 4ac)) / 2a
so by substituting,
x=(1 + sqr((-1)^2 - 4(5)(2)) / 2(5)
and
x=(1 - sqr((-1)^2 - 4(5)(2)) / 2(5)
sorry but i dont have my calculator around me, just simplify it and you'll arrive with 2 values of x. hope it helped :)
5x^2 - x + 2 = 0
now the equation is in the form of ax^2 + bx + c = 0
you can use the quadratic formula
x= (-b + sqr(b^2 - 4ac)) / 2a and
x= (-b - sqr(b^2 - 4ac)) / 2a
so by substituting,
x=(1 + sqr((-1)^2 - 4(5)(2)) / 2(5)
and
x=(1 - sqr((-1)^2 - 4(5)(2)) / 2(5)
sorry but i dont have my calculator around me, just simplify it and you'll arrive with 2 values of x. hope it helped :)
2008-11-10 9:21 am
(1+ or - sqrt(1-4(5)(2))/10
(1 + or - sqrt (1- 40))/10
since b^2<4ac, then there are no real roots.
(1 + or - sqrt (1- 40))/10
since b^2<4ac, then there are no real roots.
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