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NeX [460]
2 years ago
7

Determine a quadratic equation, in standard form, that has the pair of root: x=-3, x=5

Mathematics
2 answers:
vlabodo [156]2 years ago
6 0

Answer:

x²-2x-15=0

Step-by-step explanation:

x=-3 x=5

x+3=0 x-5=0

(x+3)(x-5)=0

x²-5x+3x-15=0

x²-2x-15=0

Mama L [17]2 years ago
5 0

Answer:

x^2 -2x -15=0

Step-by-step explanation:

\text{Given that, the roots are,}~ \alpha=-3 ~ \text{and}~ \beta=5 .\\\\\text{The quadratic equation with roots}~ \alpha ~ \text{and}~ \beta ~ \text{is,}\\\\~~~~~~~~x^2 - (\alpha + \beta )x +\alpha \beta = 0\\\\\implies x^2-(-3 +5) x +(-3)(5) = 0\\\\\implies x^2 -2x -15=0

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
I need question a and b done pls help
FrozenT [24]

Answer:

A. 5

B. 12

Step-by-step explanation:

What you do is you substitute -2 into the a and the 3 into the b.

a. 3(-2)^2-5(3)+8                       b. 2/3(3)-5(-2)

3(4)-15+8                                   2-(-10)

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3 years ago
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Help with this pleaseeee​
ELEN [110]

Answer:

(- 3, 37) and (- \frac{5}{2}, \frac{63}{2} )

Step-by-step explanation:

Given the 2 equations

2x² - y + 19 = 0 → (1)

y + 11x = 4 → (2) ← subtract 11x from both sides

y = 4 - 11x → (3)

Substitute y = 4 - 11x into (1)

2x² - (4 - 11x) + 19 = 0

2x² - 4 + 11x + 19 = 0

2x² + 11x + 15 = 0 ← in standard form

(2x + 5)(x + 3) = 0 ← in factored form

Equate each factor to zero and solve for x

2x + 5 = 0 ⇒ 2x = - 5 ⇒ x = - \frac{5}{2}

x + 3 = 0 ⇒ x = - 3

Substitute these values into (3) for corresponding values of y

x = - \frac{5}{2} : y = 4 + \frac{55}{2} = \frac{63}{2} ⇒ (- \frac{5}{2}, \frac{63}{2} )

x = - 3 : y = 4 + 33 = 37 ⇒ (- 3, 37 )

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3 years ago
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Lorico [155]

Answer:

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Step-by-step explanation:

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