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3 years ago

HELP HELP HELP

Mathematics

1 answer:

Marta_Voda [28]3 years ago

5 0

k = - 3

Other root = - 1/3

Step-by-step explanation:

$kx^2 +(5-k)x+3=0 ....(1)\\ \\ plug \: x = 3 \\ \\ k(3)^2 +(5-k) \times 3+3=0 \\ \\ 9k +15-3k+3=0 \\ \\ 6k +18=0 \\ \\ 6k = - 18 \\ \\ k = - \frac{18}{6} \\ \\ \huge \red{ k = - 3 }\\ \\ plug \: k = - 3 \: in \: equation \: (1) \\ \\ - 3 {x}^{2} + [5 - ( - 3)] x + 3 = 0 \\ \\ - 3 {x}^{2} + [5 + 3] x + 3 = 0 \\ \\ - 3 {x}^{2} + 8x + 3 = 0 \\ \\ 3 {x}^{2} - 8x - 3 = 0 \\ \\ 3 {x}^{2} - 9x + x - 3 = 0 \\ \\ 3x(x - 3) + 1(x - 3) = 0 \\ \\ (x - 3)(3x + 1) = 0 \\ \\ x - 3 = 0 \: \: or \: \: 3x + 1 = 0 \\ \\ x = 3 \: \: or \: x = - \frac{1}{3} \\ \\ \purple{ other \: root = - \frac{1}{3} }$

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If f(x) = kx^3+ x^2 − kx + 2, find a number k such that the graph of f contains the point (k, -10)

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If we write $k$ where we see $x$ in the equation and set the result equal to $-10$ , we get the result.

$f(k)=k(k)^3+k^2-k(k)+2$
$=k^4+k^2-k^2+2$
$=k^4+2=-10$
$k^4=-12$
$k_{1}=\sqrt[4]{3}(-1-i)$
$k_{2}=\sqrt[4]{3}(-1+i)$
$k_{3}=\sqrt[4]{3}(1-i)$
$k_{4}=\sqrt[4]{3}(1+i)$

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