Question

one zero of a quadratic polynomial x^2+3mx+8m is 2. find m and the other root​

Answer 1

Answer:

Step-by-step explanation:

If one of the zero is 2

Then

Putting the value of zero in the equation

x^2+3mx+8m

(2)^2+3×m×2+8m=0

4+6m+8m=0

4+14m=0

14m=-4

m=-4/14

Answer 2

Answer:

The value of 'm' = $\frac{-2}{7}$

The other root of the polynomial = $\frac{-8}{7}$

Step-by-step explanation:

Given,

One zero of the quadratic polynomial  x²+3mx+8m is 2

To find,

The value of 'm' and other root of the polynomial

Recall the theorem:

If p(x) is a polynomial and 'a' is a zero of the polynomial p(x)  then, P(a) =0

Let p(x) = x²+3mx+8m

Since it is given that '2' is a zero of the polynomial p(x) we have p(2) = 0

p(2) = 0⇒ 2²+3m×2+8m = 0

⇒ 4+6m+8m = 0

⇒ 4+14m = 0

⇒ 14m = -4

⇒ m = $\frac{-4}{14} = \frac{-2}{7}$

The value of 'm' = $\frac{-2}{7}$

Substituting the value of m in p(x) we get

p(x) =  x²+3mx+8m

= x²+3× $\frac{-2}{7}$x+8 $\frac{-2}{7}$

= $\frac{1}{7}(7x^2 -6x - 16)$

To find the other root, we need to  factorize the 7x² -6x -16

7x² -6x -16 = 7x² -14x+8x -16

= 7x(x - 2) + 8(x-2)

= (7x+8)(x-2)

7x² -6x -16 =0 ⇒ (7x+8)(x-2) = 0

7x+8 = 0 , x -2 = 0

x = $\frac{-8}{7}$ , x = 2

The other root of the polynomial = $\frac{-8}{7}$

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