How Do You Solve Quadratic Equations and Predict Zeroes Without a Calculator?

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To solve quadratic equations of the form f(x) = ax² + bx + c without a calculator, one must first convert the equation into vertex form, A(x-h)² + k. The axis of symmetry can be found using x = -b/(2a), while the vertex coordinates are (h, k), where h is derived from the axis of symmetry. The y-intercept is simply the value of c, and the number of zeroes can be predicted using the discriminant, b² - 4ac, which indicates whether there are two, one, or no real solutions. Completing the square is a key technique for transforming the equation and determining these values effectively.

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Homework Statement


f(x) : ax²+bx+c, a≠ 0

[have to find the a(x-h)²+k form first]

Find an expression in terms of a, b, c for: (without using graph or calculator)
(i) An equation of the axis of symmetry
(ii) The maximum or minimum value
(iii) The coordinated of the vertex
(iv) The domain and the range
(v) The y-intercept of the graph of the function
(vi) The zeroes of the function

Discuss how you can predict the number of zeroes for a given quadratic function of the form y= ax²+bx+c, a≠ 0. Support the validit of your prediction with some examples.



My attempt was
(ax² + bx + b/2 - b/2) + c
(ax² + bx + b/2) + c - b/2
and after this I'm stuck

(v) y intercept is (0,c)
 
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ax2+bx+c=A(x-h)2+k and expand the right side and then equate coefficients if you're unable to do it the other way.

Try adding and subtracting b/4a2 instead of b/2.
 
rock.freak667 said:
ax2+bx+c=A(x-h)2+k and expand the right side and then equate coefficients if you're unable to do it the other way.

Try adding and subtracting b/4a2 instead of b/2.

umm thanks:) but i can't make ax2+bx+c=A(x-h)2+k because ax2+bx+c is like an equation just like 3x2-6x+16
and can you please explain me how did you come up with b/4a2
 
I meant expand out A(x-h)2 and then equate 'a' to whatever the coefficient of x2 is in the expansion.

Sorry, I was supposed to type b2/4a2. Well remember that you want to have a perfect square (x-h)2 which give x2-2hx+h2 when expanded out.
 
Divina said:
umm thanks:) but i can't make ax2+bx+c=A(x-h)2+k because ax2+bx+c is like an equation just like 3x2-6x+16
and can you please explain me how did you come up with b/4a2
No, ax2+bx+c is not "like" an equation; an equation can be recognized by the presence of the = sign. ax2+bx+c is an expression, just as 3x2-6x+16 is.

Since you are NOT working with an equation, you are very limited in the things you can do. You can't add the same amount to both sides, because there are not two sides. About all you can do is add 0.

You want to complete the square by adding a certain amount and subtracting exactly the same amount so that the net change is 0.

For example,
22 + 4x
= 2(x2 + 2x)
= 2(x2 + 2x + 1) - 2
= 2(x + 1)2 - 2

In the 3rd line I added 1 inside the parentheses, but I really added 2, so to keep the expression equal to the previous one, I had to balance things by adding -2 (or subtracting 2).
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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