If the quadratic equation can be used to determine when a function equals zero, is there a modified quadratic equation that can determine when a function equals another constant?

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If the quadratic equation can be used to determine when a function equals zero, is there a modified quadratic equation that can determine when a function equals another constant?

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Answer 1 · 2015-10-24T05:33:38

$x = \frac{- b \pm \sqrt{b^{2} + 4 a (q - c)}}{2 a}$ $x = (-b+-sqrt(b^2 +4a(q-c)))/(2a)$

Explanation:

Any time you want to find for what value of $x$ $x$ a function is equal to a constant, you just set the function equal to that constant and solve for $x$ $x$ . Quadratic functions are no different. The quadratic equation is derived by setting a quadratic function equal to zero and solving for $x$ $x$ .

We can follow the original derivation but set the function to the desired constant instead of $0$ $0$ . Lets call that constant $q$ $q$ . The original derivation of the quadratic formula can be found here.

We will start by setting the general quadratic function equal to $q$ $q$ .

$a x^{2} + b x + c = q$ $ax^2 +bx+c = q$

Move $c$ $c$ to the right hand side.

$a x^{2} + b x = q - c$ $ax^2+bx=q-c$

Now we divide by $a$ $a$ .

$x^{2} + \frac{b}{a} x = \frac{q - c}{a}$ $x^2 + b/ax = (q-c)/a$

At this point we need to complete the square. Here is an explanation of what that means. We add $\frac{b^{2}}{4 a^{2}}$ $b^2/(4a^2)$ to each side in order to complete the sqare.

$x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} = \frac{q - c}{a} + \frac{b^{2}}{4 a^{2}}$ $x^2 +b/ax +b^2/(4a^2) = (q-c)/a +b^2/(4a^2)$

Now the left hand side is in a form that can be simplified into a "neat" square.

${(x + \frac{b}{2 a})}^{2} = \frac{q - c}{a} + \frac{b^{2}}{4 a^{2}}$ $(x + b/(2a))^2=(q-c)/a+b^2/(4a^2)$

At this point we want a common denominator on the right hand side in order to combine both fractions together. Multiply the first fraction by $\frac{4 a}{4 a}$ $(4a)/(4a)$ .

${(x + \frac{b}{2 a})}^{2} = \frac{4 a (q - c)}{4 a^{2}} + \frac{b^{2}}{4 a^{2}} = \frac{b^{2} + 4 a (q - c)}{4 a^{2}}$ $(x + b/(2a))^2=(4a(q-c))/(4a^2)+b^2/(4a^2)=(b^2 +4a(q-c))/(4a^2)$

I rearranged the right hand side a little bit in order to make it look more like the original quadratic equation. We can now take the square root of both sides.

$x + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2} + 4 a (q - c)}{4 a^{2}}} = \pm \frac{\sqrt{b^{2} + 4 a (q - c)}}{2 a}$ $x + b/(2a)=+-sqrt((b^2 +4a(q-c))/(4a^2))=+-sqrt(b^2 +4a(q-c))/(2a)$

Lastly, move the constant term from the left to the right by subtracting from both sides.

$x = \frac{- b \pm \sqrt{b^{2} + 4 a (q - c)}}{2 a}$ $x = (-b+-sqrt(b^2 +4a(q-c)))/(2a)$

There you have it, the only difference from the original quadratic formula is that the $- c$ $-c$ becomes $q - c$ $q-c$ . If you set $q$ $q$ equal to $0$ $0$ , you get the original quadratic formula.

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If the quadratic equation can be used to determine when a function equals zero, is there a modified quadratic equation that can determine when a function equals another constant?

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