Question

if -5 is one of the zero of 2x^2+px-15. quadratic polymial p(x^2+x)+k has both the zeroes equal to each other.find k​

Answer 1

Answer:

k=7/4

Step-by-step explanation:

first equation:

f(-5)=$2x^{2} + px-15=0$

0=2×5²-5p-15

5p=50-15

p=7

now from 2nd equation: p($x^{2} +x$)+k=0

putting the value p=7,

7$x^{2}$+7x+k=0

now, as given in the question, both zeros of the second equation is equal,

∴ discriminant,D=b²-4ac=0    (b=7, a=7 & c=k)

7²-4×7×k=0

49-28k=0

k=49/28=7/4

Hence, k=7/4

Answer 2

${ \rm{ \purple{ \underline{ \huge{QUESTION }}}}}$




▪ If -5 is one of the zero of 2x^2 + px- 15.
Quadratic polynomial p ( x^2 + x ) + k has both the zeroes equal to each other. Find k.



${ \rm{ \huge{ \purple{ \underline{SOLUTION }}}}}$



▪ Let's see the first quadratic equation..



${ \sf{ \red{2 {x}^{2} + px - 15 \: = 0}}}$


• It is given that one of the zero of the above given polynomial is equal to - 5



i. e., x = -5



Then,



putting the value of x in the equation for finding 'p'...



${ \sf{ \blue{2 {x}^{2} + px - 15 = 0}}}$



${ \implies{ \blue{ \sf{2( {-5}^{2} ) + p(-5) - 15 = 0}}}}$



${ \implies{ \blue{ \sf{(2 \times 25) - 5p - 15 = 0}}}} \\ \\ { \implies{ \blue{ \sf{50 - 15 - 5p = 0}}}} \\ \\ { \implies{ \blue{ \sf{5p - 35 = 0}}}}$




${ \implies{ \blue{ \sf{5p = 35}}}} \\ \\ { \implies{ \boxed{ \blue{ \sf{p = 7}}}}}$




▪ Now, seeing the second quadratic polynomial....



${ \red{ \sf{p( {x}^{2} + x) + k}}}$



• according to the question ,



both the zeroes of this polynomial are equal to each other....



at first putting the value of 'p' in the equation..



${ \green{ \sf{ 7( {x}^{2} + x) + k}}}$



${ \rightarrow{ \green{ \sf{ 7 {x}^{2} + 7x + k}}}}$




✒ the zeros of the quadratic polynomial ax2+bx+c,c​=0 are equal.
=> Value of the discriminant(D) has to be zero.
=>b2−4ac=0
=>b2=4ac



${ \sf{here}} \\ \\ { \sf{in \: equation}{ \pink{ \:( +7 {x}^{2} +7x + k)}}} \\ \\{ \sf{a = }{ \blue{ + 7}}} \\ \\ { \sf{b = }{ \blue{ + 7}}} \\ \\ { \sf{c = }{ \blue{k}}}$




${ \sf{since}} \\ \\ { \star{ \sf{ \red{ \: \: {b}^{2} = 4ac}}}} \\ \\ { \rightarrow{ \sf{ \green{ {( +7)}^{2} = 4( +7)k}}}}$




${ \implies{ \green{ \sf{49 = 4 \times 7 \times k}}}} \\ \\ { \implies{ \green{ \sf{k = \frac{ 49}{7 \times 4}}}}}$




${ \implies{ \underline{ \boxed{ \green{ \sf{ \: \: \: k = \frac{ 7 }{4} \: \: }}}}}}$