Quadratic Equation

Definition

The quadratic formula gives us the solutions, or roots, to a quadratic equation of the form \(ax^2 + bx + c\):

x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Technique

What is the sum of the possible solutions of x22x3=0 x^2 - 2x - 3 = 0 ?

Using the quadratic formula:

x=(2)±(2)24(1)(3)2(1)x=1±2x=3 or x=1 \begin{aligned} x &= \frac{-(-2) \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} \\ x &= 1 \pm 2 \\ x &=3 \text{ or } x=-1 \end{aligned}

The the answer is 31=2 3-1=2 . _\square

The quadratic formula is helpful even when the solutions are complex numbers:

 

The roots of x24x+5 x^2 - 4x + 5 are two complex numbers, z1 z_1 and z2 z_2 . What is the sum of z1 z_1 and z2 z_2 ?

x=(4)±(4)24(1)(5)2(1)x=4±42x=2±i \begin{aligned} x &= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)} \\ x &= \frac{4 \pm \sqrt{-4}}{2} \\ x &= 2 \pm i \end{aligned}

Thus, the two roots are z1=2i z_1 = 2-i and z2=2+i z_2 = 2 + i and their sum is (2i)+(2+i)=4 (2-i) + (2 + i ) = 4 . _\square

Whether the roots are real or complex depends on the quadratic formula's discriminant, b24ac b^2 - 4ac, the expression inside the square root. The roots are real when the discriminant is positive and complex when the discriminant is negative.

Application and Extensions

For what value of cc will 2x2+7x+c=0 2x^2 + 7x + c = 0 have only a single real root?

The quadratic formula's ±\pm tells us that there will always be two roots unless the discriminant is equal to 0. So,

b24ac=0(7)24(2)c=0c=498 \begin{aligned} b^2 - 4ac &= 0\\ (7)^2 - 4(2)c &= 0\\ c &= \tfrac{49}{8} \end{aligned} _\square

 

xx is a negative number such that x2+9x22=0 x^2+9x -22 = 0 . What is the sum of all possible values of yy which satisfy the equation x=y213y+24 x = y^2 - 13y + 24 ?

Since x2+9x22=0 x^2+9x -22 = 0 , we know that

x=9±814(22)2=9±1692=9±132=2 or 11 \begin{aligned} x &= \frac{-9 \pm \sqrt{81-4(-22)}}{2} \\ &= \frac{-9 \pm \sqrt{169}}{2} \\ &= \frac{-9 \pm 13}{2} \\ &= 2 \text{ or } -11 \end{aligned}

Since x x is negative, x=11 x=-11 . So now we need to solve 11=y213y+24 -11 = y^2 - 13y + 24 .

This produces the following quadratic equation:

y213y+35=0(y5)(y7)=0 \begin{aligned} y^2 - 13y + 35 &=0 \\ (y-5)(y-7)&=0 \end{aligned}

Thus y=5 or 7 y=5 \text{ or } 7, and the answer is 12. _\square

#Algebra #QuadraticEquations #KeyTechniques

Note by Arron Kau
7 years, 2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...