Q. \(x^2 – 11x + 19 = -5\).

Q: Solve x^2-11x+19=-5.

Convert to x^2-11x+24=0. Factor as (x-3)(x-8)=0. So x=3 or x=8.

Q: How do you rewrite the equation to standard form?

Start with x^2-11x+19=-5. Add 5 to both sides: x^2-11x+24=0.

Q: Can you factor x^2-11x+24=0 without the quadratic formula?

Need numbers multiply to 24 and add to -11. They are -3 and -8. So x^2-11x+24=(x-3)(x-8).

Q: Check the solutions in the original equation.

For x=3: 9-33+19=-5. For x=8: 64-88+19=-5. Both work.

Q: What is the discriminant and how does it confirm two solutions?

For x^2-11x+24=0, a=1,b=-11,c=24. Discriminant D=b^2-4ac=121-96=25>0, so two real solutions.

Q: What is the sum and product of the solutions?

For x^2-11x+24=0, sum =11, product =24. The solution pair 3 and 8 matches: 3+8=11, 3\cdot8=24.

Answer

Solve \(x^2-11x+19=-5\).

Bring all terms to one side:

\[
x^2-11x+24=0
\]

Factor:

\[
x^2-11x+24=(x-3)(x-8)=0
\]

So \(x=3\) or \(x=8\).

Final result: \(x=3\) or \(x=8\).

Detailed Explanation

We need to solve the equation

\[
x^2 – 11x + 19 = -5.
\]

Step 1: Move everything to one side.
To solve a polynomial equation, it is standard to have all terms on one side equal to zero. Add \(5\) to both sides:

\[
x^2 – 11x + 19 + 5 = -5 + 5.
\]

Step 2: Simplify.

\[
x^2 – 11x + 24 = 0.
\]

Step 3: Factor the quadratic.
We want to factor \(x^2 – 11x + 24\) into the form \((x-a)(x-b)\).
For \((x-a)(x-b)\), the product is \(ab = 24\) and the sum is \(a+b = 11\).

Find two numbers that multiply to \(24\) and add to \(11\).
Those numbers are \(3\) and \(8\), because \(3 \cdot 8 = 24\) and \(3 + 8 = 11\).

So the factorization is:

\[
x^2 – 11x + 24 = (x-3)(x-8).
\]

Step 4: Set each factor equal to zero.
Use the zero product property:

\[
(x-3)(x-8) = 0.
\]

Therefore, either

\[
x-3 = 0
\]

\[
x-8 = 0.
\]

Step 5: Solve each linear equation.

If \(x-3=0\), then

\[
x = 3.
\]

If \(x-8=0\), then

\[
x = 8.
\]

Final Answer:

\[
x = 3 \quad \text{or} \quad x = 8.
\]

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Algebra FAQ

Solve \(x^2-11x+19=-5\).

Convert to \(x^2-11x+24=0\). Factor as \((x-3)(x-8)=0\). So \(x=3\) or \(x=8\).

How do you rewrite the equation to standard form?

Start with \(x^2-11x+19=-5\). Add \(5\) to both sides: \(x^2-11x+24=0\).

What are the roots using the quadratic formula for \(x^2-11x+24=0\)?

\(x=\frac{11\pm\sqrt{121-96}}{2}=\frac{11\pm\sqrt{25}}{2}=\frac{11\pm5}{2}\). Thus \(x=8\) or \(x=3\).

Can you factor \(x^2-11x+24=0\) without the quadratic formula?

Need numbers multiply to \(24\) and add to \(-11\). They are \(-3\) and \(-8\). So \(x^2-11x+24=(x-3)(x-8)\).

Check the solutions in the original equation.

For \(x=3\): \(9-33+19=-5\). For \(x=8\): \(64-88+19=-5\). Both work.

What is the discriminant and how does it confirm two solutions?

For \(x^2-11x+24=0\), \(a=1,b=-11,c=24\). Discriminant \(D=b^2-4ac=121-96=25>0\), so two real solutions.

What is the sum and product of the solutions?

For \(x^2-11x+24=0\), sum \(=11\), product \(=24\). The solution pair \(3\) and \(8\) matches: \(3+8=11\), \(3\cdot8=24\).

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Q. \(x^2 – 11x + 19 = -5\).

Answer

Detailed Explanation

Algebra FAQ

Solve \(x^2-11x+19=-5\).

How do you rewrite the equation to standard form?

What are the roots using the quadratic formula for \(x^2-11x+24=0\)?

Can you factor \(x^2-11x+24=0\) without the quadratic formula?

Check the solutions in the original equation.

What is the discriminant and how does it confirm two solutions?

What is the sum and product of the solutions?

Similar Questions

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