Q. \(x^2 – 9x + 20 = 0\)

Q: Use the quadratic formula for x^2-9x+20=0.

a=1,b=-9,c=20. x=\frac{9\pm\sqrt{81-80}}{2}=\frac{9\pm1}{2}, so x=4,5.

Q: Compute the discriminant for x^2-9x+20=0.

D=b^2-4ac=81-80=1, so two distinct real roots.

Q: Find the sum and product of the roots.

Sum =9. Product =20, since for ax^2+bx+c, roots satisfy r_1+r_2=-\frac{b}{a}, r_1r_2=\frac{c}{a}.

Q: Are the roots rational, and are they unique?

The discriminant D=1 is a perfect square, so roots are rational and distinct: x=4 and x=5.

Answer

We factor the quadratic.

\[
x^2 – 9x + 20 = (x-4)(x-5)=0
\]

\[
x=4 \quad \text{or} \quad x=5
\]

Final result: \(x=4\) or \(x=5\).

Detailed Explanation

We want to solve the quadratic equation:

\[ x^2 – 9x + 20 = 0. \]

Step 1: Factor the quadratic.

To factor \(x^2 – 9x + 20\), we look for two numbers that multiply to \(20\) and add to \(-9\).

We list factor pairs of \(20\):

\(1 \cdot 20 = 20\)

\(2 \cdot 10 = 20\)

\(4 \cdot 5 = 20\)

We need the pair whose sum is \(-9\). The numbers \( -4 \) and \( -5 \) work because:

\(-4 + (-5) = -9\)

and

\((-4)(-5) = 20\)

So we can factor the quadratic as:

\[ x^2 – 9x + 20 = (x – 4)(x – 5). \]

Step 2: Use the zero product property.

If a product is zero, then at least one factor must be zero.

So:

\[ (x – 4)(x – 5) = 0. \]

This gives two cases:

\[ x – 4 = 0 \quad \text{or} \quad x – 5 = 0. \]

Step 3: Solve each equation.

Case 1:

\[ x – 4 = 0 \]

Add \(4\) to both sides:

\[ x = 4. \]

Case 2:

\[ x – 5 = 0 \]

Add \(5\) to both sides:

\[ x = 5. \]

Final Answer:

\[ x = 4 \quad \text{or} \quad x = 5. \]

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Algebra FAQ

Factor \(x^2-9x+20=0\).

\(x^2-9x+20=(x-4)(x-5)\).

Solve \(x^2-9x+20=0\) by factoring.

\((x-4)(x-5)=0\) so \(x=4\) or \(x=5\).

Use the quadratic formula for \(x^2-9x+20=0\).

\(a=1,b=-9,c=20\). \(x=\frac{9\pm\sqrt{81-80}}{2}=\frac{9\pm1}{2}\), so \(x=4,5\).

Compute the discriminant for \(x^2-9x+20=0\).

\(D=b^2-4ac=81-80=1\), so two distinct real roots.

Find the sum and product of the roots.

Sum \(=9\). Product \(=20\), since for \(ax^2+bx+c\), roots satisfy \(r_1+r_2=-\frac{b}{a}\), \(r_1r_2=\frac{c}{a}\).

Are the roots rational, and are they unique?

The discriminant \(D=1\) is a perfect square, so roots are rational and distinct: \(x=4\) and \(x=5\).

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