Write a quadratic equation with the given roots calculator

Write a Quadratic Equation Given the Roots and a Leading Coefficient

Step 1: Write the roots as factors.

Step 2: Input the factors from step 1, and the leading coefficient, into the factored form of the equation. (If you are interested in the factored form you are finished at this step!)

Step 3: Rewrite the equation from step 2 into standard form by using the distributive property and simplification.

Write a Quadratic Equation Given the Roots and a Leading Coefficient Equations

Factored form: The factored form of a quadratic equation looks like:

$$f(x) = a(x-p)(x-q) $$

• Where {eq}a {/eq} is the leading coefficient, and {eq}(x-p), (x-q) {/eq} are factors.

• We use the factors to solve for the roots as follows:

$$\begin{align} x-p &= 0&&& x-q &= 0\\ x-p +p&= 0+p&&& x-q+q &= 0+q\\ x&= p&&& x &= q \end{align} $$

• So that roots of the equation are {eq}p {/eq} and {eq}q {/eq}.

• Tip: This is important to note because we can always work backward from the roots to get the factors!

Standard form: The standard form of a quadratic equation looks like:

$$f(x) = ax^2 + bx +c $$

• Where {eq}a {/eq} is the leading coefficient.

Let's try two example problems to practice writing a quadratic equation given the roots and a leading coefficient. For the first question, we will write the equation in factored form. For the second question, we will write the equation in standard form.

Example 1: Write a Quadratic Equation Given the Roots and a Leading Coefficient: Factored Form

Find the factored form of the equation of a quadratic with roots of 6 and -12 and a leading coefficient of -7.

Step 1: Write the roots as factors.

• For this first step, we need to take the roots we've been given and rewrite them as factors. Normally when given the factored form of the equation we can pull the roots from the equation by setting the factors equal to zero and solving (as we did in the equations section above!). But we now have the roots and want the factors, so we will need to perform those same steps but in reverse!
• We have two roots as follows:

$$\begin{align} x&= 6&&& x &= -12 \end{align} $$

• Let's rewrite each of these as factors, or said another way so that they are an expression equal to zero. We do this by putting {eq}x {/eq} and the root, on the same side of the equation.

$$\begin{align} x&= 6&&& x &= -12\\ x-6&= 6-6&&& x+12 &= -12+ 12\\ x-6&= 0&&& x+12 &= 0 \end{align} $$

So, we now have two factors for our equation {eq}(x-6), (x+12) {/eq}.

Step 2: Input the factors from step 1, and the leading coefficient, into the factored form of the equation.

For the second step, we will take the factors and the leading coefficient and put it into the factored form of the equation. Keeping in mind that the factored form looks like:

$$f(x) = a(x-b)(x-c) $$

Taking the factors from step 1, and the leading coefficient of {eq}a = -7 {/eq}, we then have:

$$f(x) = -7(x-6)(x+12) $$

Therefore, the factored form of the equation of a quadratic with roots of 6 and -12 and a leading coefficient of -7 is {eq}f(x) = -7(x-6)(x+12) {/eq}.

Note: We don't need step 3 here because we want to keep the equation in the factored form!

Example 2: Write a Quadratic Equation Given the Roots and a Leading Coefficient: Standard Form

Find the standard form of the equation of a quadratic with roots of 3 and 11, and a leading coefficient of 4.

Step 1: Write the roots as factors.

We determine the factors of the equation by using the roots as we did above.

$$\begin{align} x&= 3&&& x &= 11\\ x-3&= 3-3 &&& x-11 &= 11-11\\ x-3&= 0&&& x-11 &= 0 \end{align} $$

This gives us our two factors of {eq}(x-3), (x-11) {/eq}.

Step 2: Input the factors from step 1, and the leading coefficient, into the factored form of the equation.

Inputting these two factors along with the leading coefficient of 4 we have:

$$f(x) = 4 (x-3) (x-11) $$

Step 3: Rewrite the equation from step 2 into standard form by using the distributive property and simplification.

Since the equation above is in a factored form, we need to perform the extra steps of distribution and simplification to get the equation into standard form as follows:

$$\begin{align} f(x) &= 4 (x-3) (x-11)\\ &= 4 (x^2 -11x - 3x + 33)\\ & = 4(x^2 -14x+ 33)\\ & = 4(x^2) +4(-14x)+ 4(33)\\ & = 4x^2 -56x+ 132 \end{align} $$

Therefore, the standard form of the equation of a quadratic with roots of 3 and 11 and a leading coefficient of 4 is {eq}f(x)= 4x^2 -56x+ 132 {/eq}.

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