3 years ago

Help please 19 through 23

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

Answer:

Step-by-step explanation:

hello :

19) h(x)=-2f(x)

20) p(x) = 4f(x)+1

21) m(x) = -1/3 f(x)-4 ...... continu

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What is the solution set of the equation using the quadratic formula? x2+6x+10=0 {−3+2i,−3−2i} {−6+2i,−6−2i} {−3+i,−3−i} {−2i,−4

olga2289 [7]

Answer:

The solution set of the quadratic function $x^{2}+6\cdot x +10$ is $\{-3+i,-3-i\}$ .

Step-by-step explanation:

Let be a second-order polynomial (quadratic function) is standard form and equalized to zero:

$a\cdot x^{2}+b\cdot x + c = 0$

Its roots can be determined by the Quadratic Formula in terms of its polynomial coefficients, which states that:

$x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

Given that $a = 1$ , $b = 6$ and $c = 10$ , the roots of the polynomial are, respectively:

$x_{1,2} = \frac{-6\pm \sqrt{6^{2}-4\cdot (1)\cdot (10)}}{2\cdot (1)}$

$x_{1,2} = -3\pm i$

$x_{1} = -3+i$

$x_{2} = -3 -i$

The solution set of the quadratic function $x^{2}+6\cdot x +10$ is $\{-3+i,-3-i\}$ .

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Help please 19 through 23​

Help please 19 through 23