Q. \(x^2 – 7x + 10\)

Q: Find the roots of x^2-7x+10=0 ?

Using the factorization: (x-5)(x-2)=0, so x=5 or x=2.

Q: Solve x^2-7x+10=0 by the quadratic formula ?

a=1,b=-7,c=10. x=\frac{7\pm\sqrt{49-40}}{2}=\frac{7\pm3}{2}, giving x=5,2.

Q: Compute the discriminant of x^2-7x+10 ?

\Delta=b^2-4ac=(-7)^2-4(1)(10)=49-40=9.

Q: Complete the square for x^2-7x+10 ?

x^2-7x+10=\left(x-\frac{7}{2}\right)^2-\frac{9}{4}.

Q: What is the vertex (maximum/minimum point) of y=x^2-7x+10 ?

Vertex at x=\frac{-b}{2a}=\frac{7}{2}. y=\left(\frac{7}{2}\right)^2-7\left(\frac{7}{2}\right)+10=\frac{-9}{4}.

Answer

We factor the quadratic \(x^2-7x+10\) by finding two numbers that multiply to \(10\) and add to \(-7\): \(-5\) and \(-2\).

\[\begin{aligned}
x^2-7x+10 &= (x-5)(x-2).
\end{aligned}\]

So the factored form (roots) is \(x=5\) or \(x=2\).

Detailed Explanation

We want to simplify the expression

\[x^2 – 7x + 10.\]

A common first step is to factor a quadratic when possible. We look for two numbers \(a\) and \(b\) such that:

\[a \cdot b = 10 \quad \text{and} \quad a + b = -7.\]

We test factor pairs of \(10\):

\[1 \cdot 10 = 10 \quad \text{and} \quad 1 + 10 = 11 \text{ (not } -7\text{)}.\]

\[2 \cdot 5 = 10 \quad \text{and} \quad 2 + 5 = 7 \text{ (not } -7\text{)}.\]

To get a sum of \(-7\), both numbers should be negative:

\[-2 \cdot -5 = 10 \quad \text{and} \quad -2 + (-5) = -7.\]

So the quadratic factors as:

\[x^2 – 7x + 10 = (x – 2)(x – 5).\]

Final factored form:

\[(x – 2)(x – 5).\]

See full solution

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Algebra FAQ

Factor \(x^2-7x+10\) ?

\(x^2-7x+10=(x-5)(x-2)\).

Find the roots of \(x^2-7x+10=0\) ?

Using the factorization: \((x-5)(x-2)=0\), so \(x=5\) or \(x=2\).

Solve \(x^2-7x+10=0\) by the quadratic formula ?

\(a=1,b=-7,c=10\). \(x=\frac{7\pm\sqrt{49-40}}{2}=\frac{7\pm3}{2}\), giving \(x=5,2\).

Compute the discriminant of \(x^2-7x+10\) ?

\(\Delta=b^2-4ac=(-7)^2-4(1)(10)=49-40=9\).

Complete the square for \(x^2-7x+10\) ?

\(x^2-7x+10=\left(x-\frac{7}{2}\right)^2-\frac{9}{4}\).

What is the vertex (maximum/minimum point) of \(y=x^2-7x+10\) ?

Vertex at \(x=\frac{-b}{2a}=\frac{7}{2}\). \(y=\left(\frac{7}{2}\right)^2-7\left(\frac{7}{2}\right)+10=\frac{-9}{4}\).

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