Consider the quadratic expression x^2 + 4x + c = 0. For what value of c does the equation have...

Question:

Consider the quadratic expression {eq}x^2 + 4x + c = 0 {/eq}. For what value of c does the equation have one real root?

Solutions of a Quadratic Equation:

A quadratic equation is of the form {eq}ax^2+bx+c=0 {/eq}. Its discriminant is {eq}d=b^2-4ac {/eq}. It has

(i) two real and different roots if its discriminant {eq}d>0 {/eq}.

(ii) one real root if its discriminant {eq}d=0 {/eq}.

(iii) two irrrational roots if its discriminant {eq}d<0 {/eq}.

For the roots to be rational, the discriminant bust me a perfect square.

Answer and Explanation:

The given equation is:

$$x^2 + 4x + c = 0 $$

Comparing this with {eq}ax^2+bx+c=0 {/eq}, we get:

$$a=1\\[0.4cm] b=4\\[0.4cm] c=c $$

Its discriminant is:

$$d=b^2-4ac = 4^2 -4(1)(c)=16-4c $$

For a quadratic equation to have one real root (it means one real root or two real roots), {eq}d=0 {/eq}.

Substitute the value of {eq}d {/eq} in this equation:

$$d =0 \\[0.4cm] 16-4c =0 \\[0.4cm] \text{Adding 4c on both sides},\\[0.4cm] 16 =4c \\[0.4cm] \text{Dividing both sides by 4},\\[0.4cm] 4 =c $$

Therefore, {eq}\color{blue}{\boxed{\mathbf{c=4}}} {/eq}.


Learn more about this topic:

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How to Use the Quadratic Formula to Solve a Quadratic Equation

from Math 101: College Algebra

Chapter 4 / Lesson 10
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