Question:

For what values of m can the expression $2x^2 + mxy + 3y^2 - 5y - 2$ be expressed as the product of two linear factors

Updated On: Apr 26, 2024
  • 0
  • $\pm\, 1$
  • $\pm\, 7$
  • 49
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The Correct Option is C

Solution and Explanation

We have the expression
$2x^2 + mxy + 3y^2 - 5y - 2$
Comparing the given expression with
$ax^2 + 2hxy + by^2 + 2gx + 2 fy + c$,
we get
$a= 2, h = \frac{m}{2} , b=3,c=-2 , g=0, f=- \frac{5}{2}$
The given expression is resolvable into linear factors, if
$abc + 2 fgh - af^{2} -bg^{2} -ch^{2} = 0 $
$\left(2\right)\left(3\right)\left( -2\right) + 2\left(0\right)=2 \left(\frac{25}{4}\right) - 0 -\left(-2\right) \frac{m^{2}}{4} = 0 $
$\Rightarrow - 12 - \frac{25}{2} + \frac{m^{2}}{2} = 0$
$\Rightarrow \frac{m^{2}}{2} = \frac{49}{2}$
$\Rightarrow m^{2} = 49$
$\Rightarrow m = \pm7 $
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Concepts Used:

Quadratic Equations

A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers

Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.

The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)

Two important points to keep in mind are:

  • A polynomial equation has at least one root.
  • A polynomial equation of degree ‘n’ has ‘n’ roots.

Read More: Nature of Roots of Quadratic Equation

There are basically four methods of solving quadratic equations. They are:

  1. Factoring
  2. Completing the square
  3. Using Quadratic Formula
  4. Taking the square root