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3 years ago

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Using the quadratic formula to solve x2 = 5 – x, what are the values of x? StartFraction negative 1 plus-or-minus StartRoot 21 E

ndRoot Over 2 EndFraction StartFraction negative 1 plus-or-minus StartRoot 19 EndRoot i Over 2 EndFraction StartFraction 5 plus-or-minus StartRoot 21 EndRoot Over 2 EndFraction StartFraction 1 plus-or-minus StartRoot 19 EndRoot i Over 2 EndFraction

2 answers:

nikdorinn [45]3 years ago

4 0

Answer:

Add 4

subtract 24 from 5

2

5 = –6x2 + 24x

5 = –6(x2 – 4x)

(Add 4) inside the parentheses and (subtract 24 from 5).

–19 = –6(x – 2)2

StartFraction 19 Over 6 EndFraction = (x – 2)2

Plus or minus StartRoot StartFraction 19 Over 6 EndFraction EndRoot = x – 2

The two solutions are (2)Plus or minus StartRoot StartFraction 19 Over 6 EndFraction EndRoot.

Snezhnost [94]3 years ago

3 0

Answer: StartFraction negative 1 plus-or-minus StartRoot 21 EndRoot Over 2

Step-by-step explanation:

The given quadratic equation is expressed as

x² = 5 - x

Rearranging the equation to take the standard form of ax² + bx + c, it becomes

x² + x - 5 = 0

The general formula for solving quadratic equations is expressed as

x = [- b ± √(b² - 4ac)]/2a

From the equation given,

a = 1

b = 1

c = - 5

Therefore,

x = [- 1 ± √(1² - 4 × 1 × - 5)]/2 × 1

x = [- 1 ± √(1 - - 20)]/2

x = [- 1 ± √21]/2

x = (- 1 + √21)/2 or x = (- 1 - √21)/2

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In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

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3 years ago