for the given quadratic eqation 5x + 3y = 5 state the root.
options:
(A) (1, 0)
(B) ( 2, 1)
(C) (-1, -2)
(D) ( 0, 1)
Answers
Answer:
(1) x2 + 5x – 2 = 0
(1) x2 + 5x – 2 = 0Only one variable x.
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2y
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1x
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1x
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1x =-2
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1x =-2⇒x2+1=-2x
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1x =-2⇒x2+1=-2xOnly one variable x.
(1) x2 + 5x – 2 = 0Only one variable x.Maximum index = 2(2) y2 = 5y – 10Only one variable y.Maximum index = 2So, it is a quadratic equation.(3) y2+1y =2⇒y3+1=2yOnly one variable y.Maximum index = 3So, it is not a quadratic equation.(4) x+1x =-2⇒x2+1=-2xOnly one variable x.Maximum index = 2
(5) (m + 2) (m – 5) = 0
(5) (m + 2) (m – 5) = 0⇒m2-3m-10=0
(5) (m + 2) (m – 5) = 0⇒m2-3m-10=0Only one variable m.
(5) (m + 2) (m – 5) = 0⇒m2-3m-10=0Only one variable m.Maximum index = 2
(6) m3 + 3m2 – 2 = 3m3
(6) m3 + 3m2 – 2 = 3m3Only one variable m.
(6) m3 + 3m2 – 2 = 3m3Only one variable m.Maximum index = 3
Step-by-step explanation:
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