Question

the solutions to a quadratic equation are x = 2 and x = 7. if k is a nonzero constant, which of the following must be equal to the quadratic equation?
○ ( k(x - 2)(x - 7) = 0 )
○ ( (x - 2)(x - 7) + k = 0 )
○ ( (x + 2)(x + 7) + k = 0 )
○ ( k(x + 2)(x + 7) = 0 )

Explanation:

Step1: Recall the factored form of a quadratic equation

If a quadratic equation has solutions \( x = a \) and \( x = b \), then its factored form is \( k(x - a)(x - b)=0 \), where \( k \) is a non - zero constant. This is because when \( x=a \), \( (x - a)=0 \) and when \( x = b \), \( (x - b)=0 \), and multiplying by a non - zero constant \( k \) does not change the roots (since \( k
eq0 \)).

Step2: Identify the roots and apply the factored form

The given roots of the quadratic equation are \( x = 2 \) and \( x=7 \). Using the factored form formula \( k(x - a)(x - b)=0 \) (where \( a = 2 \) and \( b=7 \)), we get the equation \( k(x - 2)(x - 7)=0 \).

Let's check the other options:

• For the option \( (x - 2)(x - 7)+k = 0 \), if we expand \( (x - 2)(x - 7)=x^{2}-9x + 14 \), then the equation is \( x^{2}-9x+(14 + k)=0 \). The roots of this equation will not be \( x = 2 \) and \( x = 7 \) in general (since adding a constant \( k\) changes the equation from the one with roots \( 2\) and \(7\)).
• For the option \( (x + 2)(x + 7)+k=0 \), expanding \( (x + 2)(x + 7)=x^{2}+9x + 14 \), the equation is \( x^{2}+9x+(14 + k)=0 \). The roots of this quadratic are not \( 2\) and \( 7\).
• For the option \( k(x + 2)(x + 7)=0 \), the roots of this equation would be \( x=-2 \) and \( x = - 7\) (when \( k

eq0\)), which is not what we want.

Answer:

A. \( k(x - 2)(x - 7)=0 \)