To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0 x = [ -b ± √(b^2 - 4ac) ] / (2a) x = [ -7 ± √((7)^2 - 4(-2)(-5)) ] / ( 2(-2) ) x = [-7 ± √(49 - (40) ) ] / ( -4 ) x = [-7 ± √(9) ] / ( -4) x = [-7 ± 3 ] / ( -4 ) x = 7/4 ± -3/4 The answers are 7/4 + 3/4 = 5/2 and 1.
First you must know this. [tex]{ax}^{2} + bx + c = 0[/tex] Then, you will know that a=-2 b=7 c=-5. Now, use the quadratic formula. [tex]x = \frac{ \: - b + - \sqrt{ {b}^{2} - 4ac } }{2a}[/tex] [tex]x = \frac{ \: - 7 + - \sqrt{ {7}^{2} - 4( - 2)( - 5) } }{2 \times - 2}[/tex] You will then get two values for x. [tex]x = 1 \: and \: x = 2.5[/tex] there you go! That's the answer.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0 x = [ -b ± √(b^2 - 4ac) ] / (2a) x = [ -4 ± √((4)^2 - 4(-3)(-5)) ] / ( 2(-3) ) x = [-4 ± √(16 - (60) ) ] / ( -6 ) x = [-4 ± √(-44) ] / ( -6) Since √-44 is nonreal, the answer to this question is that there are no real solutions.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0 x = [ -b ± √(b^2 - 4ac) ] / (2a) x = [ -5 ± √((5)^2 - 4(-11)(-3)) ] / ( 2(-11) ) x = [-5 ± √(25 - (132) ) ] / ( -22 ) x = [-5 ± √(-107) ] / ( -22) Since we conclude that √-107 is nonreal, the answer to this question is that there are no real solutions.
More analytically, we can apply the quadratic formula to find the solutions. This quadratic formula solves equations of the form ax^2 + bx + c = 0 x = [ -b ± √(b^2 - 4ac) ] / (2a) x = [ -3 ± √((3)^2 - 4(-10)(-2)) ] / ( 2(-10) ) x = [-3 ± √(9 - (80) ) ] / ( -20 ) x = [-3 ± √(-71) ] / ( -20) Since √-71 is a nonreal number, there are no real solutions to this equation.
The equation has no real solutions. (There are no x-intercepts.)
_____
The complex solutions are x = 0.625 ±i√1.359375
[tex]Solve the equation for x, where x is a real number: -4x^2 + 5x - 7 = 0[/tex]
The trestle meets ground level at 0.9 units and 9.1 units
Step-by-step explanation:
∵ -x² + 10x - 8 = 0 ⇒ × (-1)
∴ x² - 10x + 8 = 0
∵ 10x/2 = 5x ⇒ (x) × (5) ⇒ square x is x² and square 5 is 25
By using completing square form
∴ (x² - 10x + 25) - 25 + 8 = 0
∴ (x - 5)² - 17 = 0
∴ (x - 5)² = 17 ⇒ take square root for both sides
∴ x - 5 = -√17 ⇒ x = -√17 + 5 = 0.9
∴ x - 5 = √17 ⇒ x = √17 + 5 = 9.1
∴ The trestle meets ground level at x = 0.9 and x = 9.1
The equation has no real solutions. (There are no x-intercepts.)
_____
The complex solutions are x = 0.15 ±i√0.1755.
[tex]Solve the equation for x, where x is a real number: -10x^2 + 3x - 2 = 0[/tex]
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -7 ± √((7)^2 - 4(-2)(-5)) ] / ( 2(-2) )
x = [-7 ± √(49 - (40) ) ] / ( -4 )
x = [-7 ± √(9) ] / ( -4)
x = [-7 ± 3 ] / ( -4 )
x = 7/4 ± -3/4
The answers are 7/4 + 3/4 = 5/2 and 1.
The equation has no real solutions. (The graph has no x-intercepts.)
_____
The complex solutions are x = 0.4 ±i√0.44.
[tex]Solve the equation for x, where x is a real number: -5x^2 + 4x - 3 = 0[/tex]
First you must know this.
[tex]{ax}^{2} + bx + c = 0[/tex]
Then, you will know that a=-2 b=7 c=-5.
Now, use the quadratic formula.
[tex]x = \frac{ \: - b + - \sqrt{ {b}^{2} - 4ac } }{2a}[/tex]
[tex]x = \frac{ \: - 7 + - \sqrt{ {7}^{2} - 4( - 2)( - 5) } }{2 \times - 2}[/tex]
You will then get two values for x.
[tex]x = 1 \: and \: x = 2.5[/tex]
there you go! That's the answer.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -4 ± √((4)^2 - 4(-3)(-5)) ] / ( 2(-3) )
x = [-4 ± √(16 - (60) ) ] / ( -6 )
x = [-4 ± √(-44) ] / ( -6)
Since √-44 is nonreal, the answer to this question is that there are no real solutions.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -5 ± √((5)^2 - 4(-11)(-3)) ] / ( 2(-11) )
x = [-5 ± √(25 - (132) ) ] / ( -22 )
x = [-5 ± √(-107) ] / ( -22)
Since we conclude that √-107 is nonreal, the answer to this question is that there are no real solutions.
More analytically, we can apply the quadratic formula to find the solutions. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -3 ± √((3)^2 - 4(-10)(-2)) ] / ( 2(-10) )
x = [-3 ± √(9 - (80) ) ] / ( -20 )
x = [-3 ± √(-71) ] / ( -20)
Since √-71 is a nonreal number, there are no real solutions to this equation.
Theta=1.57rad, 4.71rad, 0rad, 2π rad, π rad and –π rad
[tex]Solve the equation for θ, where 0 ≤ θ ≤ 2π.[/tex]