Question

if one root of the quadratic equation 2x² +kx+6=0 is 2 then find the value of k and other root​

Answer 1

Step-by-step explanation:

given root is 2

2x²+kx+6=0

substitute x=2

2(2)²×k(2)+6=0

8+2k+6=0

2k=-14

k= -7

now substitute k=-7 in given equation

2x²-7x+6=0

2x²-3x-4x+6=0

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

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

2x-3=0 x-2=0

x=3/2 x=2

Answer 2

$\huge \orange { \underline{ \underline{ \mathfrak{Given }}}}$

• One root of the equation is 2

$\huge \orange { \underline{ \underline{ \mathfrak{To\: Find}}}}$

• Value of k
• the second root

$\huge \orange { \underline{ \underline{ \mathfrak{Solution}}}}$

Let the name of equation be p(x)

$\red{\sf{So,}}$

$\ratio \implies$p(x) = 2x² + kx + 6 =0

$\red{\sf{Putting\: value\: of\: first\: root}}$

2(2)² + k(2) + 6 = 0 ...(since we are putting the value of a root then the value of whole eqⁿ must be 0)

8 + 2k + 6 = 0

14 + 2k = 0

2k = -14

$\ratio \implies$ k = -7

$\large\orange{\sf{So\: the\: value\: of\: k = -7,}}$

Since we now know the value of k we will put it in the eqⁿ and factories it

$\ratio \implies$ p(x) = 2x² -7x + 6 = 0

$\red{\sf{Factorising \: using\: middle\: term\: split\: method}}$

2x² - 7x + 6 = 0

2x² - 4x - 3x + 6 = 0

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

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

$\red{\sf{Since\: we\: the\: first\: root\: we:will\: directly \: find \: the\: value\: of\: second\: root}}$

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

$\ratio \implies$ x = 2 or 3/2

$\large\orange{\sf{Hence,\:the\: value\: of\: second\: root\: is\: 3/2}}$

Hence, the value of :-

• k = -7
• second root = 3/2

Thank you !