Solve each of the following quadratic equations-9 - 2-9 a-b x-2 a 2-5 a b-2 b 2-0 - CBSE 2009-

We write, 9a+bx= 32a+bx 3a+2bx as 9x2 #215;2a2+5ab+2b2=92a2+5ab+2b2x2= 32a+bx #215; 3a+2bx #8756;9x2 9a+bx+2a2+5ab+2b2=0 #8658;9x2 32a+bx 3a+2bx+2a+ba ...

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Standard X

Mathematics

Elimination Method of Finding Solution of a Pair of Linear Equations

Solve each of...

Question

Solve each of the following quadratic equations:

$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$ [CBSE 2009]

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Solution

We write, $- 9 (a + b) x = - 3 (2 a + b) x - 3 (a + 2 b) x$ as $9 x^{2} \times (2 a^{2} + 5 a b + 2 b^{2}) = 9 (2 a^{2} + 5 a b + 2 b^{2}) x^{2} = [- 3 (2 a + b) x] \times [- 3 (a + 2 b) x]$
$∴ 9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0 \Rightarrow 9 x^{2} - 3 (2 a + b) x - 3 (a + 2 b) x + (2 a + b) (a + 2 b) = 0 \Rightarrow 3 x [3 x - (2 a + b)] - (a + 2 b) [3 x - (2 a + b)] = 0 \Rightarrow [3 x - (2 a + b)] [3 x - (a + 2 b)] = 0$
$\Rightarrow 3 x - (2 a + b) = 0 or 3 x - (a + 2 b) = 0 \Rightarrow x = \frac{2 a + b}{3} or x = \frac{a + 2 b}{3}$
Hence, $\frac{2 a + b}{3}$ and $\frac{a + 2 b}{3}$ are the roots of the given equation.

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Solve each of the following quadratic equations: 9x2-9a+bx+2a2+5ab+2b2=0 [CBSE 2009]

We write, -9a+bx=-32a+bx-3a+2bx as 9x2×2a2+5ab+2b2=92a2+5ab+2b2x2=-32a+bx×-3a+2bx ∴9x2-9a+bx+2a2+5ab+2b2=0⇒9x2-32a+bx-3a+2bx+2a+ba+2b=0⇒3x3x-2a+b-a+2b3x-2a+b=0⇒3x-2a+b3x-a+2b=0 ⇒3x-2a+b=0 or 3x-a+2b=0⇒x=2a+b3 or x=a+2b3 Hence, 2a+b3 and a+2b3 are the roots of the given equation.

Solve each of the following quadratic equations:

$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$ [CBSE 2009]