Question

If -5 is one of the zeroes of 2x^2+px-15.quadratic polynomial p(x^2+x)+k has both zeroes equal to one another then find k.

Answer 1

Answer:

Step-by-step explanation:

first solve fro $2x^2+px-15$

given -5 is its zero that means

$2(-5)^2+p(-5)-15=0$

after solving this you will get p=-13

put in next one you get

$-13(x^2+x)+k$    =$-13x^2-13x+k$  it has both zero equal the apply rule

discrement is zero i.e $b^2-4ac=0$

Answer 2

Appropriate Question:

$\sf{If \: -5 \: is \: one \: of \: the \: zeroe s \: o f \: 2x^2+px-15 \: quadratic \: polynomial } \\ \sf{ p(x^2+x) \: +k has \: both \: zeroes \: equal \: to \: one \: another \: then \: find \: k.}$

Solution:

$\sf{Here, - 5 \: is \: a \: zero \: of \: p(x)= {2x}^{2} + px - 15.}$

$\sf{Then, \: p( - 5) = 0 }$

$: \implies \sf{2 {( - 5)}^{2} + p + 35 = 0 } \\ : \implies \sf{2(25) - 5 - 15 = 0} \\ : \implies \sf{50 - 5p - 15 = 0} \\: \implies \sf{ - 5p + 35 = 0} \\ : \implies \sf{ - 5p = - 35} \\: \implies \underline{\sf{ \fbox{p = 7}}} \sf{..(i)}$

$\sf{We \: Have, \: f(x) = p( {x}^{2} + x) + k = 7( {x}^{2} + x) + k} \: \: \: ... [From \: (i)] \\ : \implies \sf{f(x)=p(x) = 7 {x }^{2} + 7x + k.}$

$\sf{It \: is \: given \: that \: both \: the \: zeroes \: of \: f(x) \: are \: equal, i.e., \alpha = \beta .}$

Then,

$\sf{ \alpha + \beta = \frac{ - 7}{7} } \\ : \implies \sf{ \alpha + \alpha = - 1} \\ : \implies \sf{2 \alpha = - 1} \\ : \implies \sf{ \alpha = \frac{ - 1}{2} }$

Also,

$: \implies \sf{ \alpha \beta = \frac{k}{7} } \\ : \implies \sf{ \frac{ - 1}{2} \times \bigg[ \frac{-1}{2} \bigg] = \frac{k}{7} } \\ : \implies \sf{ \frac{1}{4} = \frac{k}{7} } \\ : \implies \sf{k = \frac{7}{7} }$

$\large \mathfrak{Hence, \: k = \frac{7}{4} .}$

Final Answer:

$\LARGE\mathfrak{k = \frac{7}{4} }$