<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.
Radical form for 112 would be 4{7 and then I think you can just search up how to reduce the 4 to 7 i don’t have the symbol btw
2x^2+3. the 2x cancel out and the rest is combining like terms
First you minus 2y on both sides and you get: 5x=-2y+4. Then you minus 4 on both sides to get 5x-4=-2y which in turn is y (-2y) = mx (5x) + b (-4) : -2y=5x-4
