Find the roots of the equation $2 x^{2} - 5 x + 3 = 0$ , by factorization.

$- \frac{3}{2}, 1$

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$4, 2$

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$- \frac{5}{2}, \frac{3}{2}$

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$\frac{3}{2}, 1$

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Solution

The correct option is D
$\frac{3}{2}, 1$

We need to find two numbers whose product is $(3 \times 2) = 6$ and whose sum is $- 5$ .
Let us first split the middle term $- 5 x$ as $- 5 x = (- 2 x) + (- 3 x)$
$[∴ (- 2 x) \times (- 3 x) = 6 x^{2} = (2 x^{2}) \times 3]$ .

So, $2 x^{2} - 5 x + 3 = 2 x^{2} - 2 x - 3 x + 3 = 2 x (x - 1) - 3 (x - 1) = (2 x - 3) (x - 1)$

Now, $2 x^{2} - 5 x + 3 = 0$ can be rewritten as $(2 x - 3) (x - 1) = 0$ .
$\Rightarrow 2 x - 3 = 0$ or $x - 1 = 0$ .
Now, $2 x - 3 = 0$ gives $x = \frac{3}{2}$
and $x - 1 = 0$ gives $x = 1$ .
So, $\frac{3}{2}$ and 1 are the roots of the quadratic equation $2 x^{2} - 5 x + 3 = 0.$

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