Finding the only multiple zero of a quadratic function

Question

I don't know how to figure this problem out. Can someone please help?

For what value of $\;b\;$ will $\;f(x)=x^2 + bx + 400\;$ have $\;-20\;$ as its only zero?

Answer 1

The number $\;-20\;$ is a zero of the function $\;f(x)=x^2+bx+400\;$ if and only if $\;f(-20)=0\;.$

$(-20)^2+b(-20)+400=0\;,$

$400-20b+400=0\;,$

$-20b=-800\;,$

$20b=800\;,$

$b=\dfrac{800}{20}=40\;.$

Moreover, if $\;b=40\;,\;$ then it results that $\;f(x)=x^2+40x+400=(x+20)^2\;,$

consequently,

$f(x)=(x+20)^2=0\iff x+20=0\iff x=-20\;,$

so $\;-20\;$ is the only zero of $\;f(x)\;.$

Hence, the value of $\;b\;$ such that $\;f(x)=x^2+bx+400\;$ has $\;-20\;$ as its only zero, is

$b=40\;.$

Another way to solve the exercise:

$f(x)=x^2+bx+400\;.$

$x^2+bx+400=0\;,$

and, by applying the quadratic formula’s determinant,

$\Delta=b^2-4ac=b^2-4\cdot1\cdot400=b^2-1600\;.$

$f(x)=x^2+bx+400\;$ has only a zero if and only if $\;\Delta=0\;.$

$b^2-1600=0\;,$

$b^2=1600\;,$

$b=\pm\sqrt{1600}=\pm40\;.$

If $\;b=-40\;,\;x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{40\pm0}{2}=20\;.$

If $\;b=40\;,\;x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-40\pm0}{2}=-20\;.$

Hence, the value of $\;b\;$ such that $\;f(x)=x^2+bx+400\;$ has $\;-20\;$ as its only zero, is

$b=40\;.$

Answer 2

For a quadratic equation having one root, its discriminant is zero, i.e.

$$b^2- 4\cdot 400=0$$

which leads to $b=\pm40$. Check to determine that $b=40$ from the given root $-20$.