Well, without even looking at it, I can say it's a downwards facing parabola, which roots at x = -3 and x = 5. That should do it.
<u>We'll assume the quadratic equation has real coefficients</u>
Answer:
<em>The other solution is x=1-8</em><em>i</em><em>.</em>
Step-by-step explanation:
<u>The Complex Conjugate Root Theorem</u>
if P(x) is a polynomial in x with <em>real coefficients</em>, and a + bi is a root of P(x) with a and b real numbers, then its complex conjugate a − bi is also a root of P(x).
The question does not specify if the quadratic equation has real coefficients, but we will assume that.
Given x=1+8i is one solution of the equation, the complex conjugate root theorem guarantees that the other solution must be x=1-8i.
Substitute y = 3x + 15 to the equation -4x + 7y = 20:
-4x + 7(3x + 15) = 20 <em>use distributive property</em>
-4x + (7)(3x) + (7)(15) = 20
-4x + 21x + 105 = 20 <em>subtract 105 from both sides</em>
17x = -85 <em>divide both sides by 17</em>
x = -5
Substitute the value of x to the equation y = 3x + 15:
y = 3(-5) + 15
y = -15 + 15
y = 0
<h3>Answer: x = -5 and y = 0</h3>
9 serving
hope this helps
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