Question

solve value of x :√(2x+9) +x-13=0
please show solution.

Answer 1

hey sakna here

√2x+9  +  x             =              13

√2x+9                     =              13-x

squaring both sides

2x + 9                        =              (13-x)²

2x+9                          =              169 - 26x + x²

2x +26x +9 - 169       =              x²

28x-160                     =               x²

by using factorization method

x²-28x+160                =                0

x² - 20x - 8x +160      =                0

x(x-20) -8(x-20)          =                0

(x-20)(x-8)                   =                0

x = 20 and x= 8

hope it help

plz mark as brainliest

Answer 2

The solutions of the equation $\sqrt{(2x+9)}=-x+13$ are 20 and 8.

Given:

An equation √(2x+9) +x-13=0.

To Find:

The solution of the given equation.

Solution:

Shift the terms that are not in square root to the right-hand side of the equation.

$\sqrt{(2x+9)}=-x+13$

Square both sides of the equation.

$(\sqrt{(2x+9)})^2=(-x+13)^2\\2x+9=(x^2+169-26x)$

Combine the like terms.

$x^2-28x+160=0$

Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Substitute 1 for a, -28 for b, and 160 for c into $x=\frac{-b- \sqrt{b^2-4ac}}{2a}$ and find the first root.

$x=\frac{-(-28)- \sqrt{(-28)^2-4\cdot 1\cdot 160}}{2\cdot 1}\\=\frac{28- \sqrt{784-640}}{2}\\=\frac{28- \sqrt{144}}{2}\\=\frac{28- 12}{2}\\=\frac{16}{2}\\=8$

Substitute 1 for a, -28 for b, and 160 for c into $x=\frac{-b+ \sqrt{b^2-4ac}}{2a}$ and find the first root.

$x=\frac{-(-28)+ \sqrt{(-28)^2-4\cdot 1\cdot 160}}{2\cdot 1}\\=\frac{28+ \sqrt{784-640}}{2}\\=\frac{28+ \sqrt{144}}{2}\\=\frac{28+ 12}{2}\\=\frac{40}{2}\\ =20$

Thus, the roots of the equation $\sqrt{(2x+9)}=-x+13$ are 20 and 8.