Question

If the sum of the zeros of p(x)=(k-3)x^2-12x+5 is 4, then find the value of k.​

Answer 1

Step-by-step explanation:

sum of zeroes =-b/a=4

b=-12,a=k-3

-(-12)/k-3=4

4(k-3)=12

4k-12=12

4k=12+12

4k=24

k=24/4

k=3

Answer 2

${ \sf{ \huge \underline{solution}}}$

${ \rm{ \large {the \: value \: of \: x = 4}}}$

${ \rm{ \large {(k - 3) {x}^{2} - 12x + 5 = 0}}}$

${ \rm{ \large { \implies(k - 3) {4}^{2} - 12 \times 4 + 5 = 0}}}$

${ \rm{ \large { \implies16(k - 3) - 48 + 5 = 0}}}$

${ \rm{ \large { \implies 16k - 48 - 48 + 5 = 0}}}$

${ \rm{ \large { \implies 16k -93+ 5 = 0}}}$

${ \rm{ \large { \implies 16k -88 = 0}}}$

${ \rm{ \large { \implies 16k = 88}}}$

${ \rm{ \large { \implies k = \dfrac{88}{16} }}}$

${ \rm{ \large { \implies k = 5.5}}}$

${ \rm{ \large { \therefore \: the \: value \: of \: k = 5.5}}}$