The difference between the roots of the quadratic equation x2 -14x +q =0 is 6. Find q
1 answer:
9514 1404 393
Answer:
q = 40
Step-by-step explanation:
When the quadratic has roots p and r, it can be factored as ...
(x -p)(x -r) = x² -(p+r)x +pr
So, the sum of the roots is 14, and their difference is 6. This lets us find the roots from ...
p + r = 14
p - r = 6
2p = 20 . . . add the two equations
p = 10
r = 14 -p = 4
The value of interest is then ...
q = pr = (10)(4)
q = 40
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The graph shows the roots to be 4 and 10, as we found.
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The residuals of the linear regression equation are 0.329, -0.346, 0.064, -0.326, 0.049, -0.286, -0.161, 0.214, 0.429 and 0.059
<h3>How to determine the residuals?</h3>The regression equation is given as:
y = 0.005x + 3.111
Next, we calculate the predicted values (y) at the corresponding x values.
So, we have:
y = 0.005 * 4 + 3.111 = 3.131
y = 0.005 * 3 + 3.111 = 3.126
y = 0.005 * 5 + 3.111 = 3.136
y = 0.005 * 3 + 3.111 = 3.126
y = 0.005 * 8 + 3.111 = 3.151
y = 0.005 * 15 + 3.111 = 3.186
y = 0.005 * 10 + 3.111 = 3.161
y = 0.005 * 15 + 3.111 = 3.186
y = 0.005 * 6 + 3.111 = 3.141
y = 0.005 * 6 + 3.111 = 3.141
The residuals are then calculated using:
Residual = Actual value - Predicted value
So, we have:
y = 3.46 - 3.131 = 0.329
y = 2.78 - 3.126 = -0.346
y = 3.2 - 3.136 = 0.064
y = 2.8 - 3.126 = -0.326
y = 3.2 - 3.151 = 0.049
y = 2.9 - 3.186 = -0.286
y = 3 - 3.161 = --0.161
y = 3.4 - 3.186 = 0.214
y = 3.57 - 3.141 = 0.429
y = 3.2 - 3.141 = 0.059
Hence, the residuals of the linear regression equation are 0.329, -0.346, 0.064, -0.326, 0.049, -0.286, -0.161, 0.214, 0.429 and 0.059
See attachment for the residual plot.
The residual plot shows that the linear model from the regression calculator is a good model because the points are not on a straight line
Read more about residuals at:
#SPJ1
