Q. \(x^{2} + 2x – 15 = 0\)

Q: How do I factor x^2+2x-15?

Find numbers with product -15 and sum 2: 3 and -5. So x^2+2x-15=(x+5)(x-3).

Q: Solve x^2+2x-15=0 by factoring.

Set (x+5)(x-3)=0. Then x+5=0 \Rightarrow x=-5, or x-3=0 \Rightarrow x=3.

Q: Use the quadratic formula.

For ax^2+bx+c=0, x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. Here x=\dfrac{-2\pm\sqrt{4+60}}{2}=\dfrac{-2\pm 8}{2}, giving x=3 or x=-5.

Q: What is the discriminant of x^2+2x-15=0?

a=1,b=2,c=-15. Discriminant D=b^2-4ac=4-4(1)(-15)=4+60=64. Since D>0, there are two real solutions.

Q: How can I check the solutions quickly?

Substitute. For x=3: 9+6-15=0. For x=-5: 25-10-15=0. Both work.

Q: What are the sum and product of the roots?

From x^2+2x-15=0. Sum =-\dfrac{b}{a}=-2. Product =\dfrac{c}{a}=-15. Roots 3 and -5 match: 3+(-5)=-2, 3(-5)=-15.

Answer

We solve \(x^2+2x-15=0\) by factoring. Find numbers that multiply to \(-15\) and add to \(2\): \(5\) and \(-3\).

\[
x^2+2x-15=(x+5)(x-3)=0
\]

So \(x+5=0\) or \(x-3=0\).

\[
x=-5 \quad \text{or} \quad x=3
\]

Final answer: \(x=-5, 3\).

Detailed Explanation

We want to solve the equation \(x^2+2x-15=0\).

Step 1: Factor the quadratic (when possible)

To factor \(x^2+2x-15\), we look for two numbers that satisfy:

Their product is \(-15\)
They add up to \(2\)

The numbers \(-3\) and \(5\) work because:

\((-3)\cdot 5=-15\)
\((-3)+5=2\)

So the quadratic factors as:

\[
x^2+2x-15=(x-3)(x+5)
\]

Step 2: Use the zero product property

If \((x-3)(x+5)=0\), then at least one factor must be zero.

So we set each factor equal to zero:

\[
x-3=0
\]
\[
x+5=0
\]

Step 3: Solve each simple equation

First equation:

\[
x-3=0
\]

Add \(3\) to both sides:

\[
x=3
\]

Second equation:

\[
x+5=0
\]

Subtract \(5\) from both sides:

\[
x=-5
\]

Final Answer

The solutions to \(x^2+2x-15=0\) are:

\[
x=3 \quad \text{or} \quad x=-5
\]

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Algebra FAQ

How do I factor \(x^2+2x-15\)?

Find numbers with product \(-15\) and sum \(2\): \(3\) and \(-5\). So \(x^2+2x-15=(x+5)(x-3)\).

Solve \(x^2+2x-15=0\) by factoring.

Set \((x+5)(x-3)=0\). Then \(x+5=0 \Rightarrow x=-5\), or \(x-3=0 \Rightarrow x=3\).

Use the quadratic formula.

For \(ax^2+bx+c=0\), \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). Here \(x=\dfrac{-2\pm\sqrt{4+60}}{2}=\dfrac{-2\pm 8}{2}\), giving \(x=3\) or \(x=-5\).

What is the discriminant of \(x^2+2x-15=0\)?

\(a=1,b=2,c=-15\). Discriminant \(D=b^2-4ac=4-4(1)(-15)=4+60=64\). Since \(D>0\), there are two real solutions.

How can I check the solutions quickly?

Substitute. For \(x=3\): \(9+6-15=0\). For \(x=-5\): \(25-10-15=0\). Both work.

What are the sum and product of the roots?

From \(x^2+2x-15=0\). Sum \(=-\dfrac{b}{a}=-2\). Product \(=\dfrac{c}{a}=-15\). Roots \(3\) and \(-5\) match: \(3+(-5)=-2\), \(3(-5)=-15\).

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