Question

By using the method of completing the square, show that the equation 4x² +3x+5=0 has no real roots.

Answer 1

Answer:

Refer to the above attachment ⬆️

Answer 2

Solution :

We have a quadratic equation:

$\bullet$ $\huge{\boxed{\sf{4x^{2} - 3x + 5 = 0}}}$

So, first we write it as:

$\implies$ $\sf{4x^{2} - 3x = - 5}$

Now we divide whole equation by coefficient of x², so we get:

$\implies$ $\sf{x^{2} - 34x = - 54}$

Now to find out the constant term, which should be added and subtracted to form a complete square, we take coefficient of x - term.

So,

We add and subtract 9/64, and get:

$\implies$ $\sf{{x}^{2} - \frac{3}{4} x + \frac{9}{64} - \frac{9}{64} = - \frac{5}{4}}$

$\implies$ $\sf{ {x}^{2} - \frac{3}{4} x + \frac{9}{64} = - \frac{5}{4} + \frac{9}{64}}$

$\implies$ $\sf{(x - \frac{3}{8} ) ^{2} = \frac{ - 80 + 9}{64}}$

$\implies$ $\sf{(x - \frac{3}{8} ) = \pm \sqrt{ \frac{ - 71}{64} }}$

$\implies$ $\sf{x = \frac{3}{8} \pm \sqrt{ \frac{ - 71}{64} }}$

Here,

We have $\pm \sqrt{ \frac{ - 71}{64} }$ that has a negative number $\sf{(-71)}$ inside the square root, so we can say that the given quadratic equation has no real roots.

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Answered by: Niki Swar, Goa❤️