Question

The difference between the roots of the quadratic equation x2−14x+q=0 is 6. Find q.

Answer 1

Answer:

Step-by-step explanation:

In a quadratic equation, a rule known as Vieta's Theorem tells us that roots 1 and 2 (we'll call them x1 and x2) added up is equal to -b/a, and x1 multiplied by x2 is equal to c/a.  because x1-x2 is 6, and x1+x2=-b/a, which is -(-14), or 14, we can use the substitution method to find out the roots of the equation, being 4 and 10. When putting these roots into the equation, we get that 4^2-14(4)+q=0, and 10^2-14(10)+q=0. We can simplify these equations to make -40+q=0, and adding 40 on both sides gets us that q=40.

To recap:

Vieta's Formula:

• root 1 + root 2 = -b/a

• root 1 * root 2 = c/a

use substitution to find the roots

enter the roots into the quadratic equation

solve the equation!

Answer: q = 40

Answer 2

Answer:

$q=40$

Step-by-step explanation:

1)To answer this question we need to remember the Vieta's or Viete's Formula to find the the roots of a quadratic expression:

$ax^{2}+bx+c=0$

Since x' and x'' are the roots of this equation, then we can write the Vieta's formula:

$\left\{\begin{matrix}x'+x''=\frac{-b}{a} & \\ x'*x''=\frac{c}{a} & \end{matrix}\right.$

2) As the question gives a precious information. x'-x''=6, let's replace the 2nd equation of the Vieta's formula by that and make it a Linear System.

$\left\{\begin{matrix}x'+x''=\frac{14}{1} & \\ x'-x''=6 & \end{matrix}\right.\Rightarrow 2x'=20\therefore x'=10$

3) Solving the system:

$10+x''=14\Rightarrow x''=14-10\Rightarrow x''=4$

4) Since we have a=1:

$x^{2}-14x+q=0$

Then we can make q as the parameter c.

$q=x'*x''=4*10\therefore q=40$

$x^{2}-14x+40=0$