Question

Find the roots of the following quadratic equations, if they exist, by the method
completing the square:
(1) 2x-7x+3=0
(ii) 2x+x-4=0
please solve this question​

Answer 1

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Question:

Find the roots of the following quadratic equations, if they exist, by the method - completing the square:

(i) 2x² - 7x + 3 = 0

(ii) 2x² + x - 4 = 0

Solution:

1) 2x² - 7x + 3 = 0

We divide the whole equation by 2 so that the coefficient of x² is 1 and then transpose the constant part to the RHS so that we can move on further steps.

$\implies x^2 - \dfrac{7}{2}x + \dfrac{3}{2} = 0 \\\\\implies x^2 - \dfrac{7}{2}x = -\dfrac{3}{2}$

We can also write the above equation as,

$\implies x^2 - 2 * \dfrac{7}{4}*x = -\dfrac{3}{2}$

Hence, we add (7/4)² on both sides.

$\implies x^2 - 2 * \dfrac{7}{4}*x + \left(\dfrac{7}{4}\right)^2  = -\dfrac{3}{2} + \left(\dfrac{7}{4}\right)^2$

We know that a² - 2ab + b² = (a - b)².

So,

$\implies \left(x - \dfrac{7}{4}\right)^2 = \dfrac{49}{16} - \dfrac{3}{2} \\ \\\implies \left(x - \dfrac{7}{4}\right)^2 = \dfrac{49}{16} - \dfrac{24}{16} \\\\\implies \left(x - \dfrac{7}{4}\right)^2 = \dfrac{25}{16}$

Square-rooting both sides,

$\implies x - \dfrac{7}{4} = \pm \dfrac{5}{4} \\\\\implies x - \dfrac{7}{4} = \dfrac{5}{4} \:\:\:\:OR\:\:\:\:x - \dfrac{7}{4} = - \dfrac{5}{4} \\\\\implies x = \dfrac{5}{4} + \dfrac{7}{4} \:\:\:\:OR\:\:\:\:x = - \dfrac{5}{4} + \dfrac{7}{4} \\\\\bf{\implies x = \dfrac{12}{4} = 3\:\:\:\:OR\:\:\:\:x = \dfrac{2}{4} = \dfrac{1}{2} }$

2) 2x² + x - 4 = 0

We divide the whole equation by 2 so that the coefficient of x² is 1 and then transpose the constant part to the RHS so that we can move on further steps.

$\implies x^2 + \dfrac{1}{2}x - 2 = 0 \\\\\implies x^2 + \dfrac{1}{2}x = 2$

We can also write the above equation as,

$\implies x^2 + 2 * \dfrac{1}{4}*x = 2$

Hence, we add (1/4)² on both sides.

$\implies x^2 + 2 * \dfrac{1}{4}*x + \left(\dfrac{1}{4}\right)^2  = 2 + \left(\dfrac{1}{4}\right)^2$

We know that a² + 2ab + b² = (a + b)².

So,

$\implies \left(x + \dfrac{1}{4}\right)^2 = 2 + \dfrac{1}{16} \\ \\\implies \left(x + \dfrac{1}{4}\right)^2 = \dfrac{32}{16} + \dfrac{1}{16} \\\\\implies \left(x + \dfrac{1}{4}\right)^2 = \dfrac{33}{16}$

Square-rooting both sides,

$\implies x + \dfrac{1}{4} = \pm \dfrac{\sqrt{33}}{4} \\\\\implies x = \dfrac{\sqrt{33}}{4} - \dfrac{1}{4} \:\:\:\:OR\:\:\:\: x = -\dfrac{\sqrt{33}}{4} - \dfrac{1}{4}\\\\\bf{\implies x = \dfrac{\sqrt{33}-1}{4} \:\:\:\:OR\:\:\:\: x = -\dfrac{\sqrt{33}-1}{4}}$

Answer 2

Answer:

1. x = { 3, 1/2 }

2. x = { 5/4, -(7/4) }

Step-by-step explanation:

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