- #1
DaalChawal
- 85
- 0
Are You Struggling with Quadratic Equations?
- Context: MHB
- Thread starter DaalChawal
- Start date
Click For Summary
SUMMARY
The discussion focuses on solving quadratic equations through the manipulation of coefficients derived from roots. It establishes that for the equations represented as $(x-\alpha)(x-\beta)+(x-\gamma)(x-\delta)=0$ and $(x-m)(x-n)=0$, the relationships between the roots and coefficients can be expressed through specific equations. The key conclusion is that the roots $x_1$ and $x_2$ can be determined by substituting the relationships of $m$, $n$, $\alpha$, $\beta$, $\gamma$, and $\delta$ into the derived equations, ultimately leading to the correct identification of the roots.
PREREQUISITES- Understanding of quadratic equations and their standard forms
- Familiarity with polynomial factorization techniques
- Knowledge of Vieta's formulas relating roots and coefficients
- Basic algebraic manipulation skills
- Study Vieta's formulas in depth to understand the relationship between roots and coefficients
- Practice solving quadratic equations using different methods, including factoring and completing the square
- Explore advanced topics in polynomial equations, such as the Rational Root Theorem
- Investigate the graphical representation of quadratic functions and their roots
Students, educators, and anyone interested in mastering quadratic equations and their applications in algebra and calculus.
- #2
I like Serena
Science Advisor
Homework Helper
MHB
- 16,335
- 258
Hi DaalChawal,
We can write
$$(x-\alpha)(x-\beta)+(x-\gamma)(x-\delta)=2x^2-(\alpha+\beta+\gamma+\delta)x+(\alpha\beta+\gamma\delta)=0\implies $$
$$x^2-\frac 12(\alpha+\beta+\gamma+\delta)x+\frac 12(\alpha\beta+\gamma\delta) = 0\tag 1$$
Since $m$ and $n$ are roots, we must have
$$(x-m)(x-n)=0\implies x^2-(m+n)x+mn=0\tag 2$$
The coefficients of equations (1) and (2) must be the same, so we have:
$$\begin{cases}m+n=\frac 12(\alpha+\beta+\gamma+\delta)\\mn=\frac 12(\alpha\beta+\gamma\delta)\end{cases}\tag 3$$
We have
$$2(x-m)(x-n)-(x-\alpha)(x-\beta)=x^2-(2(m+n)-\alpha-\beta)x+(2mn-\alpha\beta)=0$$
Let its roots be $x_1$ and $x_2$, then we must have:
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$
Again, the coefficients must match, so we have
$$\begin{cases}x_1+x_2=2(m+n)-\alpha-\beta \\ x_1x_2=2mn-\alpha\beta\end{cases}\tag 4$$
Substitute (3) in (4) to find:
$$\begin{cases}x_1+x_2=2\cdot\frac 12(\alpha+\beta+\gamma+\delta)-\alpha-\beta = \gamma+\delta \\
x_1x_2=2\cdot \frac 12(\alpha\beta+\gamma\delta)-\alpha\beta = \gamma\delta\end{cases}\tag 5$$
Therefore (B) is the correct answer.
We can write
$$(x-\alpha)(x-\beta)+(x-\gamma)(x-\delta)=2x^2-(\alpha+\beta+\gamma+\delta)x+(\alpha\beta+\gamma\delta)=0\implies $$
$$x^2-\frac 12(\alpha+\beta+\gamma+\delta)x+\frac 12(\alpha\beta+\gamma\delta) = 0\tag 1$$
Since $m$ and $n$ are roots, we must have
$$(x-m)(x-n)=0\implies x^2-(m+n)x+mn=0\tag 2$$
The coefficients of equations (1) and (2) must be the same, so we have:
$$\begin{cases}m+n=\frac 12(\alpha+\beta+\gamma+\delta)\\mn=\frac 12(\alpha\beta+\gamma\delta)\end{cases}\tag 3$$
We have
$$2(x-m)(x-n)-(x-\alpha)(x-\beta)=x^2-(2(m+n)-\alpha-\beta)x+(2mn-\alpha\beta)=0$$
Let its roots be $x_1$ and $x_2$, then we must have:
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$
Again, the coefficients must match, so we have
$$\begin{cases}x_1+x_2=2(m+n)-\alpha-\beta \\ x_1x_2=2mn-\alpha\beta\end{cases}\tag 4$$
Substitute (3) in (4) to find:
$$\begin{cases}x_1+x_2=2\cdot\frac 12(\alpha+\beta+\gamma+\delta)-\alpha-\beta = \gamma+\delta \\
x_1x_2=2\cdot \frac 12(\alpha\beta+\gamma\delta)-\alpha\beta = \gamma\delta\end{cases}\tag 5$$
Therefore (B) is the correct answer.
- #3
DaalChawal
- 85
- 0
Thank you Sir
Similar threads
High School
Strange quadratic manipulation
- · Replies 6 ·
- · Replies 8 ·
- · Replies 2 ·
- · Replies 2 ·
High School
Is this curve quadratic or exponential?
- · Replies 3 ·
- · Replies 16 ·
- · Replies 6 ·
High School
Complete the square, yes, but what does it mean?
- · Replies 19 ·
- · Replies 1 ·
- · Replies 4 ·