Misaka solved the radical equation x – 3 = square root of 4x-7 but did not check her solutions. (x – 3)2 = square root of 4x-7^2 x2 – 6x + 9 = 4x – 7 x2 – 10x + 16 = 0 (x – 2)(x – 8) = 0 x = 2 and x = 8 Which shows the true solution(s) to the radical equation x – 3 = square root of 4x-7 x = 2 x = 8 x = 2 and x = 8 There are no true solutions to the equation.

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calculista calculista · Answer 1 · 2019-08-04T02:15:14+02:00

Answer:

x=8 is a true solution of the radical equation

Step-by-step explanation:

we have

[tex]x-3=\sqrt{4x-7}[/tex]

Solve for x

squared both sides

[tex](x-3)^{2}=4x-7\\\\x^{2}-6x+9=4x-7\\\\ x^{2}-10x+16=0[/tex]

Convert to factored form

[tex]x^{2}-10x+16=(x-2)(x-8)[/tex]

The solutions are x=2 and x=8

Verify the solutions

For x=2

Substitute in the original equation

[tex]2-3=\sqrt{4(2)-7}[/tex]

[tex]-1=1[/tex] ----> is not true

therefore

x=2 is not a true solution of the radical equation

For x=8

Substitute in the original equation

[tex]8-3=\sqrt{4(8)-7}[/tex]

[tex]5=5[/tex] ----> is true

therefore

x=8 is a true solution of the radical equation