Question

(x+7)^2=x^2+kx+49........k=???​

Answer 1

Step-by-step explanation:

The given quadratic equation is x

The given quadratic equation is x 2

The given quadratic equation is x 2 −kx+9=0, comparing it with ax

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k)

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2 −36≥0

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2 −36≥0⇒ k

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2 −36≥0⇒ k 2

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2 −36≥0⇒ k 2 ≥36

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2 −36≥0⇒ k 2 ≥36⇒ k≥6 or k≤−6

The given quadratic equation is x 2 −kx+9=0, comparing it with ax 2 +bx+c=0∴ We get, a=1b=−k,c=9⇒ It is given that roots are real and distinct.∴ b 2 −4ac≥0⇒ (−k) 2 −4(1)(9)≥0⇒ k 2 −36≥0⇒ k 2 ≥36⇒ k≥6 or k≤−6∴ We can see values of k given in question are correct.

Answer 2

Answer:

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Step-by-step explanation:

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