Question

Q4. Solve for x: √(2x+9) +x=13

Answer 1

Answer:

2x+9+x²=169

x²+2x-160=0

solve by this

formula method u get urs answer perfectly

Answer 2

Answer:

$\boxed{x = 8}$

Step-by-step explanation:

$Solve \: for \: x \: over \: the \: real \: numbers: \\ = > \sqrt{2x + 9} + x = 13 \\ \\ Subtract \: x \: from \: both \: sides: \\ = > \sqrt{2x + 9} + (x - x) =13-x \\ = > \sqrt{2x + 9} = 13 - x \\ \\ Raise \: both \: sides \: to \: the \: power \: of \: two: \\ = > {( \sqrt{2x + 9} )}^{2} = {(13 - x)}^{2} \\ = > {(2x + 9)}^{ \cancel{2} \times \frac{1}{ \cancel{2}} } = {(13 - x)}^{2} \\ = > 2x + 9 = {(13 - x)}^{2} \\ \\ Expand \: out \: terms \: of \: the \: right \: hand \: side: \\ = > 2x+9=( {13}^{2} - 2(13)(x) + {x}^{2} ) \\ = > 2x + 9 = 169 - 26x + {x}^{2}$ $\\ Subtract \: {x}^{2} - 26 x + 169 \: from \: both \: sides: \\ = > (2x + 9) - ( {x}^{2} - 26x + 169) = 169 - 26x + {x}^{2} - ( {x}^{2} - 26x + 169) \\ = > 2x + 9 - {x}^{2} + 26x - 169 = \cancel{169} + \cancel{- 26x} + \cancel{{x}^{2}} + \cancel{- {x}^{2} } + \cancel{ 26x} + \cancel{- 169} \\ = > - {x}^{2} + 28x - 160 = 0 \\ \\ The \: left \: hand \: side \: factors \: into \: a \: product \: with \: three \: terms: \\ = > - ( {x}^{2} - 28x + 160) \\ = > - ( {x}^{2} - (20 + 8)x + 160) \\ = > - ( {x}^{2} - 20x - 8x + 160) \\ = > - (x(x - 20) - 8(x - 20)) \\ = > - (x - 20)(x - 8)$ $\\Multiply \: both \: sides \: by - 1: \\ = > - (x - 20)(x - 8) \times ( - 1) = 0 \times ( - 1) \\ = > (x - 20)(x - 8) = 0 \\ \\ Split \: into \: two \: equations: \\ = > x - 20=0 \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: x-8=0 \\ \\ Add \: 20 \: to \: both \: sides: \\ = > \boxed{x=20} \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: x-8=0 \\ \\ Add \: 8 \: to \: both \: sides: \\ = > x = 20 \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \boxed{x = 8}$ $\\ when \: x = 8 \\ = > \sqrt{2x + 9} + x \\ = > \sqrt{2 \times 8 + 9} + 8 \\ = > \sqrt{25} + 8 \\ = > 5 + 8 \\ = > 14 = 13 : \\ So \: this \: solution \: is \: correct \\ \\ \\ when \: x = 20 \\ = > \sqrt{2x + 9} + x \\ = > \sqrt{2 \times 20 + 9} + 20 \\ = > \sqrt{49} + 20 \\ = > 7 + 20 \\ = > 27 \neq 13: \\ So \: this \: solution \: is \: incorrect \\ \\ \\ The \: solution \: is: \\ x = 8$