Question

$x {}^{2} + 4 \sqrt{3x} - 15$
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Answer 1

Step-by-step explanation:

$\bf \huge \underline \purple{Question:-}$

$\huge \sf \: {x}^{2} + 4 \sqrt{3} x - 15$

$\huge \sf \underline \purple{Answer:-}$

$\sf \longmapsto \large \: {x}^{2} + 4 \sqrt{3} x - 15 = 0$

$\sf \longmapsto \large \: {x}^{2} + 5 \sqrt{3} x - \sqrt{3} x - 15 = 0$

$\sf \longmapsto \large \:x(x + 5 \sqrt{3} ) - \sqrt{3} (x + 5 \sqrt{3} ) = 0$

$\sf \longmapsto \large \:(x - \sqrt{3} )(x + 5 \sqrt{3} ) = 0$

$\sf \longmapsto \large \:x - \sqrt{3} = 0 \: \: and \: \: x + 5 \sqrt{3} = 0$

$\sf \longmapsto \large \:x = \sqrt{3} \: \: and \: \: x = - 5 \sqrt{3}$

$\sf \large \: x \: should \: \: not \: be \: \: in \: \: negative$

$\therefore \sf \large \blue { x = \sqrt{3} }$

Answer 2

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.