C-05-Q167mediumsingle_mcq

The substitution method for $\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0$ uses—

acompleting square
bletting $y = \dfrac{x + a}{x - a}$ to reduce to quadratic in $y$
csquaring both sides
dcross-multiplication only

ব্যাখ্যা

Setting $y=\dfrac{x+a}{x-a}$ converts the expression into the simple quadratic $y^2-5y+6=0$ . Solving for $y$ and then back-substituting recovers $x$ .

এই অধ্যায়ের আরও প্রশ্ন

Equations in One Variable — সব প্রশ্ন দেখুন (170টি) →

The substitution method for $\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0$ uses—

acompleting square

bletting

y = \dfrac{x + a}{x - a}

to reduce to quadratic in

y

csquaring both sides

dcross-multiplication only

ব্যাখ্যা

Setting $y=\dfrac{x+a}{x-a}$ converts the expression into the simple quadratic $y^2-5y+6=0$ . Solving for $y$ and then back-substituting recovers $x$ .

The substitution method for (x+ax−a)2−5(x+ax−a)+6=0\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0(x−ax+a​)2−5(x−ax+a​)+6=0 uses—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The substitution method for (x+ax−a)2−5(x+ax−a)+6=0\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0(x−ax+a​)2−5(x−ax+a​)+6=0 uses—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The substitution method for $\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0$ uses—

The substitution method for $\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0$ uses—