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Stells [14]
3 years ago
6

Solve the equation: w^2+7w+12=0

Mathematics
1 answer:
polet [3.4K]3 years ago
7 0
Hi there!
We are given the equation w² + 7w + 12 = 0, and we are told to solve it. Well, we can first take all the factors of 12 -
1 12
2 6
3  4
Now, take the sum of each factor pair -
1, 12 = 13
2, 6 = 8
3, 4 = 7
Find which factor pair adds up to 7, and we can see that 3 and 4 add up to seven, while also having a product of 12. Therefore, since the whole equation has addition signs, we can factor the equation w² + 7w + 12 into (w + 3)(w + 4) = 0. Next, using the Zero Product Property, we can set each term to zero.
w + 3 = 0
w = -3

w + 4 = 0
w = -4
Therefore, the solution to the equation w² + 7w + 12 = 0 is w = -3, -4. Hope this helped and have a great day!
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One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

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So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

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Cross Multiply

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kolbaska11 [484]

Answer:

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Step-by-step explanation:

There are a couple of ways to go at these. One is to use the sum and difference formulas for the cosine and sine functions. To do that, you need to find the sine for the angle whose cosine is given, and vice versa.

Another approach is to use the inverse trig functions to find the angles α and β, then combine those angles and find find the desired function of the combination.

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