Question

If x = 4 is one root of (k + 2) x2 – (5k + 2)x – 4 = 0, find the value of k. Also,
find the other root of the equation.

Answer 1

Got $x=4$ as a root?

Steps : Substitution → Find $k$ → Find the other root → END

Substitution → Find $k$

$16(k+2)-4(5k+2)-4=0$

$16k+32-20k-8-4=0$

$-4k=-20$

∴ $k=5$

Find $k$ → Find the other root

$7x^2-27x-4=0$

$(7x+1)(x-4)=0$

∴ $x=-\frac{1}{7} ,4$

Find the other root → END

The other root is $-\frac{1}{7}$.

Answer 2

Answer:

The value of k = 5

The other root of the equation =  $\frac{-1}{7}$

Step-by-step explanation:

Given,

x = 4 is one root of (k + 2) x² – (5k + 2)x – 4 = 0

To find,

The value of 'k' and the other root of the equation

Solution:

Recall the concept:

If x = a, is a root of the polynomial p(x), then p(a) = 0

Let p(x) =  (k + 2) x² – (5k + 2)x – 4 = 0

Since x = 4, is a root of the polynomial p(x) we have p(4) = 0

p(4) = 0 ⇒  (k + 2) ×4² – (5k + 2)×4 – 4  = 0

⇒  (k + 2) ×16 – (5k + 2)×4 – 4  = 0

⇒  16k+32 - 20k - 8 - 4  = 0

⇒  -4k+20  = 0

⇒  -4k = -20

⇒  k = 5

∴ The value of k = 5

Substituting the value of k, in p(x) we get

(5+2) x² – (5×5 + 2)x – 4 = 0

7x² – 27x – 4 = 0

7x² – 28x + x– 4 = 0

7(x –  4) +1(x– 4) = 0

(7x+1)(x– 4) = 0

(7x+1) = 0, (x– 4) = 0

x = $\frac{-1}{7}$ and x = 4

∴The other root of the equation =  $\frac{-1}{7}$

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