Q. Solve \(x^2 – 6x = 16\) by completing the square.

Q: What are the solutions of x^2-6x=16 by completing the square?

Rewrite as x^2-6x+9=16+9, so (x-3)^2=25. Then x-3=\pm5, giving x=8 or x=-2.

Q: Why do we add 9 when completing the square?

For x^2-6x take half the linear coefficient: (-6/2)^2=9. Adding 9 produces a perfect square trinomial (x-3)^2; you must add the same 9 to the other side to keep the equation balanced.

Q: What if completing the square yields a negative on the right side?

If you get (x-h)^2=k with k<0, there are no real solutions; solutions are complex: x=h \pm i\sqrt{|k|}. Completing the square still works and gives complex roots when k<0.

Q: How does completing the square relate to the vertex form of a parabola?

Completing the square converts ax^2+bx+c into a(x-h)^2+k, the vertex form. The vertex is (h,k). For x^2-6x, you get (x-3)^2-9, so the vertex is (3,-9).

Q: Can completing the square derive the quadratic formula?

Can completing the square derive the quadratic formula?

Q: How can I check my solutions quickly?

Substitute each solution back into the original equation to verify equality. For x=8: 64-48=16. For x=-2: 4+12=16. Both satisfy the equation.

Q: Common mistakes when completing the square?

Frequent errors: forgetting to add the same value to both sides, using b^2 instead of (b/2)^2, sign mistakes when taking square roots, or failing to divide by a when a\neq1.

Answer

\[
\begin{aligned}
x^2-6x &= 16 \\
x^2-6x+9 &= 16+9 \quad\text{(add }(6/2)^2=9\text{)} \\
(x-3)^2 &= 25 \\
x-3 &= \pm 5 \\
x &= 3\pm 5 \\
x &= 8 \text{ or } x = -2
\end{aligned}
\]

Detailed Explanation

Write the original equation.

\[x^{2}-6x=16\]
Identify the coefficient of x and compute half of it. The coefficient of x is -6, so half is -3. Then square that half to find the number to add to both sides. Half squared is

\[(-3)^{2}=9\]
Add 9 to both sides of the equation so the left side becomes a perfect square trinomial. This preserves equality because the same number is added to each side.

\[x^{2}-6x+9=16+9\]
Factor the perfect square trinomial on the left and simplify the right side. The left factors as (x minus 3) squared because the middle term is twice the product of x and -3, and the constant term is (-3) squared.

\[(x-3)^{2}=25\]
Take the square root of both sides. Remember to include both the positive and negative square roots.

\[x-3=\pm 5\]
Solve the two linear equations obtained from the plus and minus cases separately.

Case 1: \[x-3=5\] which gives \[x=8\].

Case 2: \[x-3=-5\] which gives \[x=-2\].
Check both solutions by substitution into the original equation.

For x = 8: \[8^{2}-6(8)=64-48=16\] which is true.

For x = -2: \[(-2)^{2}-6(-2)=4+12=16\] which is true.
Therefore, the solutions are

\[x=8\quad\text{or}\quad x=-2\]

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FAQs

What are the solutions of \(x^2-6x=16\) by completing the square?

Rewrite as \(x^2-6x+9=16+9\), so \((x-3)^2=25\). Then \(x-3=\pm5\), giving \(x=8\) or \(x=-2\).

Why do we add \(9\) when completing the square?

For \(x^2-6x\) take half the linear coefficient: \((-6/2)^2=9\). Adding \(9\) produces a perfect square trinomial \((x-3)^2\); you must add the same \(9\) to the other side to keep the equation balanced.

How do you complete the square when the leading coefficient is not 1?

First divide the entire equation by the leading coefficient \(a\). For \(ax^2+bx+c=0\) get \(x^2+(b/a)x+\dots\). Then add \((b/(2a))^2\) to both sides to form a perfect square \((x+b/(2a))^2\).

What if completing the square yields a negative on the right side?

If you get \((x-h)^2=k\) with \(k<0\), there are no real solutions; solutions are complex: \(x=h \pm i\sqrt{|k|}\). Completing the square still works and gives complex roots when \(k<0\).

How does completing the square relate to the vertex form of a parabola?

Completing the square converts \(ax^2+bx+c\) into \(a(x-h)^2+k\), the vertex form. The vertex is \((h,k)\). For \(x^2-6x\), you get \((x-3)^2-9\), so the vertex is \((3,-9)\).

Can completing the square derive the quadratic formula?

How can I check my solutions quickly?

Substitute each solution back into the original equation to verify equality. For \(x=8\): \(64-48=16\). For \(x=-2\): \(4+12=16\). Both satisfy the equation.

Common mistakes when completing the square?

Frequent errors: forgetting to add the same value to both sides, using \(b^2\) instead of \((b/2)^2\), sign mistakes when taking square roots, or failing to divide by \(a\) when \(a\neq1\).

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Q. Solve \(x^2 – 6x = 16\) by completing the square.

Answer

Detailed Explanation

FAQs

What are the solutions of \(x^2-6x=16\) by completing the square?

Why do we add \(9\) when completing the square?

How do you complete the square when the leading coefficient is not 1?

What if completing the square yields a negative on the right side?

How does completing the square relate to the vertex form of a parabola?

Can completing the square derive the quadratic formula?

How can I check my solutions quickly?

Common mistakes when completing the square?

Similar Questions

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