Question

when
$\sqrt{2x + 1} + \sqrt{3x + 4} = 7$
the value of x is
​

Answer 1

GIVEN :-

• √(2x + 1 ) + √(3x + 4) = 7

TO FIND :-

• The value of x.

SOLUTION :-

→ √(2x+1) + √(3x+4) = 7

→ √(2x+1) = 7 - √(3x+4)

★ Squaring on both sides ★

→ 2x + 1 = 49 + 3x + 4 - 14√(3x + 4)

→ 2x + 1 = 53 + 3x - 14√(3x+4)

→ 14√(3x+4) = x + 52

★ Again squaring on both sides ★

→ 14²(3x+4) = x² + 52² + 104x

→ 196(3x+4) = x² + 2704 + 104x

→ 588x + 784 = x² + 2704 + 104x

→ x² - 484 x + 1920 = 0

→ x² - 4x - 480x + 1920 = 0

→ x ( x - 4 ) - 480 ( x - 4 ) = 0

→ ( x - 4 ) ( x - 480 ) = 0

→ x = 4 & x = 480

Hence the required value of x is 4 and 480 when √(2x + 1 ) + √(3x + 4) = 7.

Answer 2

Given :-

$\to\sf\sqrt{2x + 1} + \sqrt{3x + 4} = 7$

To Find :-

$\to\sf Value \: of \: x$

Solution :-

$:\implies$ √(2x + 1) + √(3x + 4) = 7

$:\implies$ √(2x + 1) = 7 - √(3x + 4)

★ Squaring on both sides ★

$:\implies$ 2x + 1 = 49 + 3x + 4 - 14√(3x + 4)

$:\implies$ 2x + 1 = 53 + 3x - 14√(3x + 4)

$:\implies$ 14√(3x + 4) = x + 52

★ Squaring on both sides ★

$:\implies$ 14² (3x + 4) = x² + 52² + 104x

$:\implies$196 (3x + 4) = x² + 2704 + 104x

$:\implies$ 588x + 784 = x² + 2704 + 104x

$:\implies$ x² - 484x + 1920 = 0

$:\implies$ x (x - 4) - 480 (x - 4) = 0

$:\implies$ (x - 4) (x - 480) = 0

$:\implies$ x = 4 & x = 480

Hence, the value of x is 4 & 480 when √(2x + 1) + √(3x + 4) = 7.