MHB Are You Struggling with Quadratic Equations?

AI Thread Summary
The discussion focuses on solving quadratic equations by relating the roots and coefficients of two quadratic expressions. It establishes that if \( m \) and \( n \) are roots, their relationships can be expressed through equations derived from the coefficients of the quadratics. By equating coefficients from two sets of equations, the roots \( x_1 \) and \( x_2 \) are determined to be \( \gamma \) and \( \delta \). The conclusion confirms that the correct answer is (B), demonstrating a clear method for solving quadratic equations through systematic substitution and comparison. This approach effectively illustrates the relationships between roots and coefficients in quadratic equations.
DaalChawal
Messages
85
Reaction score
0
1628607224461.png
 
Mathematics news on Phys.org
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.
 
Thank you Sir🙂
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top