Question

if the equation x^2-5x+6=0 and x^2-7x+k=0 have a common root ,then the value of k is​

Answer 1

Answer:

k = 10 or k = 12

Step-by-step explanation:

x² - 5x + 6 = 0 and x² - 7x + k = 0

Let's consider

x² - 5x + 6 = 0

ax² + bx + c = 0

where a = 1, b = -5, c = 6

By splitting the middle term method,

Sum = b = -5

Product = a × c = 6

Thus factors are -3 and -2

x² - 3x - 2x + 6 =0

x(x - 3) -2(x - 3) = 0

(x - 2)(x - 3) = 0

So, x = 2 or x = 3

If x = 2 is the common root then,

p(x) = x² - 7x + k

thus,

p(2) = 2² - 7(2) + k

0 = 4 - 14 + k

0 = -10 + k

therefore, k = 10

OR

If the common root is x = 3 then

p(3) = 3² - 7(3) + k

0 = 9 - 21 + k

0 = -12 + k

k = 12

so, k = 10 or k = 12 when the common root is x = 2 or x = 3

respectively

That is,

If the common root is x = 2 then k = 10

If the common root is x = 3 then k = 12

Hope you understood it........All the best

Answer 2

Answer:

Value of k is 12 or 10.

Step-by-step explanation:

Given two equations are

${x}^{2} - 5x + 6 = 0........(1) \: \: and \: \: {x}^{2} - 7x + k = 0........(2)$

For the equation no. (1),

${x}^{2} - 5x + 6 = 0$

By middle term method we can solve it,

${x}^{2} - (3 + 2)x + 6 = 0$

${x}^{2} - 3x - 2x + 6 = 0$

$x( x - 3) - 2(x - 3) = 0$

$(x - 3)(x - 2) = 0$

If multiplication of two term is 0 then they are separately 0.

$(x - 3) = 0 \: \: \: or \: \: \: ( x - 2) = 0$

So

$x = 3 \: \: \: or \: \: \: x = 2$

Now according to question equation number (2) has same root of equation number (1).

So we are putting x=3 and 2 in the equation (2) and get,

${3}^{2} - 7 \times 3 + k = 0 \\ 9 - 21 + k = 0 \\ - 12 + k = 0 \\ k = 12$

and

${2}^{2} - 7 \times 2 + k = 0 \\ 4 - 14 + k = 0 \\ k = 10$

For x=3, value of k is 12

and for x=2,value of k is 10.