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3 years ago

Consider the quadratic equation

Mathematics

2 answers:

maria [59]3 years ago

6 0

Answer:

A: -19

B: The equation has two complex solutions.

Step-by-step explanation:

(A) Compare

... x² +5x +11 = 0

to the form

... ax² + bx + c = 0

and you see that a=1, b=5, c=11.

The discriminat (d) is computed as

... d = b²-4ac

Putting the above values in this equation for a, b, c, we get

... d = 5² -4·1·11 = 25 -44 = -19

_____

(B) The solutions are ...

... x = (-b ±√d)/(2a) = (-5 ±√-19)/2

The square root of -19 is imaginary, so there are two complex solutions. It will be the case that the two solutions are complex whenever the discriminant is negative.

Sergeeva-Olga [200]3 years ago

5 0

$\bf \qquad \qquad \qquad \textit{discriminant of a quadratic} \\\\\\ \stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+5}x\stackrel{\stackrel{c}{\downarrow }}{+11}=0 ~~~~~~~~ \stackrel{discriminant}{b^2-4ac}= \begin{cases} 0&\textit{one real solution}\\ positive&\textit{two real solutions}\\ negative&\textit{no real solution}\\ &\boxed{\textit{two complex solutions}} \end{cases} \\\\\\ (5)^2-4(1)(11)\implies 25-44\implies -19\impliedby \boxed{negative}$

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Solve the system of equations: y = x + 2 y = x2 + 5x + 6 HELP PLZ

soldier1979 [14.2K]

Answer:

x=2(twice)

y=0(twice)

Step-by-step explanation:

This question can be solved using substitution method

So let's solve

y=x+2....(1)

y=x2+5x+6....(2)

Substitute (1) into(2)

X+2=x2+5x+6

Collect like terms

X2+5x-x+6-2=0

X2+4x+4=0

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(X+2)(x+2)=0

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Substrate 2 from both sides

X=-2

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4 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ (Approximated)

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

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Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ (Approximated)

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

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