Answered

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Find one value of x that is a solution to the equation:(4x – 1)^2 = 20x – 5

Sagot :

To find the value of x, you can follow the steps.

Step 1: Solve the quadratic part of the equation.

Knowing that

[tex](a+b)^2=a^2+2ab+b^2[/tex]

Then,

[tex]\begin{gathered} (4x-1)^2=20x-5 \\ 16x^2-8x+1=20x-5 \end{gathered}[/tex]

Step 2: Add - 20x +5 to each side of the equation to write the quadratic equation in the standard formula.

[tex]\begin{gathered} 16x^2-8x+1-20x+5=20x-5-20x+5 \\ 16x^2-28x+6=0 \end{gathered}[/tex]

Use the Bhaskara formula to find the roots of the equation.

For a quadratic equation ax² + bx + c = 0, the Bhaskara for is:

[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]

Step 3: Substitute the constants and find x.

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