Question

By the method of completion of squares show that the equation 4x² + 3x + 5 = 0 has no real roots.​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {4x {}^{2} + 3x + 5 = 0}$

$\implies \mathsf {x {}^{2} + \frac{3}{4}x + \frac{5}{4} = 0}$

$\implies \mathsf {x {}^{2} + \frac{3}{4}x + ( \frac{3}{8}) {}^{2} = \frac{ - 5}{4} + \frac{9}{64}}$

$\implies \mathsf {(x + \frac{3}{8}) {}^{2} = \frac{ - 71}{64}}$

$\implies \mathsf {x + \frac{3}{8} = \sqrt{ \frac{ - 71}{64} } \: not \: a \: real \: number.}$

$\mathsf {Hence, \: QE \: has \: no \: real \: roots.}$

Answer 2

Completion Of Square

• First factor out 4, except the constant term.

$4(x^2+\frac{3}{4} x)+5$

• Then square '(linear constant)/2' and add it inside.

$4(x^2+\frac{3}{4}x +\frac{9}{64} -\frac{9}{64} )+5$

• Take negative out.

$4(x^2+\frac{3}{4}x +\frac{9}{64} )+5-\frac{9}{16}$

• Take the L.C.M, Completion Of Square

$4(x+\frac{3}{8} )^2+\frac{71}{16}$

_______________________

$4(x+\frac{3}{8} )^2+\frac{71}{16}=0$

• Isolate constant term.

$4(x+\frac{3}{8} )^2=-\frac{71}{16}$

• Divide both sides.

$(x+\frac{3}{8} )^2=-\frac{71}{64}$

According to the definition of the square root

If you square $x+\frac{3}{8}$ you get $-\frac{71}{64}$.

Only imaginary number gives negative after squaring.

No Real Root

Thank you for reading such a long answer.