Write a quadratic equation in the variable X. Write in standard form: ax^2+bx+c=0 using 3 as the only solution.
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If there is only one root, it occurs twice:
(x-3)(x-3) = 0
x^2 - 6x + 9 = 0
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The solution is x^2-6x+9=0? In ax^2 + bx + c =0 form? Can you explain please?
To write a quadratic equation with 3 as the only solution, we need to factorize it based on the given solution. Since 3 is the only solution, it means that (x - 3) is a factor of the quadratic equation.
To find the other factor, we divide the quadratic equation by (x - 3). We use the long division method to find the quotient.
Let's assume our original quadratic equation is ax^2 + bx + c. Dividing it by (x - 3), we get:
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(x - 3) | ax^2 + bx + c
In the long division process, we start by dividing the highest degree term of the dividend (ax^2) by the highest degree term of the divisor (x), which gives us a.
a
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(x - 3) | ax^2 + bx + c
Next, we multiply the divisor (x - 3) by a and subtract it from the dividend to find the next term. When we multiply a by (x - 3), we get ax - 3a. Subtracting this term from the dividend, we get:
ax^2 + bx + c - (ax - 3a)
= bx + c + 3a
Now, we repeat the process by considering the highest degree term of the updated dividend (bx). We divide bx by x, which gives us b as the next term.
a + b
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(x - 3) | ax^2 + bx + c
Next, just like before, we multiply the divisor (x - 3) by (a + b) and subtract it from the updated dividend to find the remainder.
bx + c + 3a - (bx - 3(a+b))
= c + 6a - 3b
Now, we have the remainder c + 6a - 3b. Since (x - 3) is a factor, the remainder must be zero for 3 to be the only solution. Therefore, we have the equation:
c + 6a - 3b = 0
This equation is in standard form (ax^2 + bx + c = 0). Thus, the quadratic equation with 3 as the only solution is:
(c + 6a - 3b)x^2 + bx + c = 0