Question

${ \sqrt{8x + 13 \: \: } } + { \sqrt{2x - 3}} \: \: \: \: equals \: to \: 5 \\ \\find \: the \: \:value \: of \: x \: and \: prove \: it$
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Answer 1

Answer:

$\sqrt{(8x + 13)} + \sqrt{(2x - 3)} = 5 \\ \\ on \: \: squaring \: \: both \: \: sides \\ = > 8x + 13 + 2x - 3 + 2 \sqrt{6x + 10} = 5 \\ = > 6x + 10 + 2 \sqrt{6x + 10} = 5 \\ = > 2 \sqrt{6x + 10} = - 6x - 5 \\ = > \sqrt{6x + 10} = \frac{ - 6x - 5}{2} \\ \\ on \: squaring \: both \: sides \\ = > 6x + 10 = (36 {x}^{2} + 60x + 25) \div 4 \\ = > 24x + 40 = 36 {x}^{2} + 60x + 25 \\ = > 36 {x}^{2} + 36x - 15 = 0 \\ = > 12 {x}^{2} + 12x - 5 = 0 \\ on \: \: application \: \: of \\ quadratic \: \: formula \\ x = \frac{ - 12( + - ) \sqrt{384} }{24}$

Answer 2

Answer:

Hyy there

Answer:

\begin{gathered} \sqrt{(8x + 13)} + \sqrt{(2x - 3)} = 5 \\ \\ on \: \: squaring \: \: both \: \: sides \\ = > 8x + 13 + 2x - 3 + 2 \sqrt{6x + 10} = 5 \\ = > 6x + 10 + 2 \sqrt{6x + 10} = 5 \\ = > 2 \sqrt{6x + 10} = - 6x - 5 \\ = > \sqrt{6x + 10} = \frac{ - 6x - 5}{2} \\ \\ on \: squaring \: both \: sides \\ = > 6x + 10 = (36 {x}^{2} + 60x + 25) \div 4 \\ = > 24x + 40 = 36 {x}^{2} + 60x + 25 \\ = > 36 {x}^{2} + 36x - 15 = 0 \\ = > 12 {x}^{2} + 12x - 5 = 0 \\ on \: \: application \: \: of \\ quadratic \: \: formula \\ x = \frac{ - 12( + - ) \sqrt{384} }{24} \end{gathered}

(8x+13)

+

(2x−3)

=5

onsquaringbothsides

=>8x+13+2x−3+2

6x+10

=5

=>6x+10+2

6x+10

=5

=>2

6x+10

=−6x−5

=>

6x+10

=

2

−6x−5

onsquaringbothsides

=>6x+10=(36x

2

+60x+25)÷4

=>24x+40=36x

2

+60x+25

=>36x

2

+36x−15=0

=>12x

2

+12x−5=0

onapplicationof

quadraticformula

x=

24

−12(+−)

384