Question

Solve the following quadratic equations by completing the square method. (a) x²-6x +4=0​

Answer 1

Answer:

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Answer 2

Answer:

Force a perfect square trinomial on the left side of the equation. Take the square root of both sides and solve for x

$x {}^{2} + 6x - 4 = 0$

Force a perfect square trinomial such that

$a {}^{2} + 2ab + b {}^{2} = (a + b) {}^{2}$

on the left side of the equation. Then take the square root of both sides and solve for x

Add 4 to both sides of the equation.

$x {}^{2} + 6 x = 4$

Take the coefficient of the x term and divide it by 2

Take the coefficient of the x term and divide it by 2 and square the result.

$\frac{6}{2} = 3 \: \: 3 { }^{2} = 9$

Add 9 to both sides.

${x}^{2} + 6x + 9 = 4 + 9$

$= x {}^{2} + 6x + 9 = 13$

$a = x \ \: : b = 3$

$(x + 3)3 {}^{2} = 13$

Take the square root of both sides.

$x + 3 = ± \sqrt{13}$

Subtract 3 from both sides.

$x = - 3 + \sqrt{13}$

$x = - 3 - \sqrt{13}$