Q. Write the quadratic equation in standard form: \(5x^2 + 5x = 1\).

Q: How do you write the equation in standard form?

Move all terms to one side so the equation equals zero: 5x^2 + 5x - 1 = 0.

Q: How do I know this is a quadratic?

A quadratic has the form ax^2 + bx + c with a \ne 0. Here a = 5, b = 5, c = -1, so it is quadratic.

Q: Can the equation be simplified by factoring out a common factor?.

No integer common factor > 1 exists (gcd5,5,-1=1). Dividing by 5 gives fractions: x^2 + x - 1/5 = 0..

Q: How do I find the roots using the quadratic formula?

Use x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=5,b=5,c=-1. Discriminant 45, so x = \frac{-5 \pm \sqrt{45}}{10} = \frac{-5 \pm 3\sqrt{5}}{10}..

Q: How many real solutions does the equation have?

Discriminant \Delta = 45 > 0 , so there are two distinct real roots.

Q: What is the vertex and axis of symmetry?

Axis: x = -b/(2a) = -5/10 = -1/2. Vertex: plug in to get y = 5(-1/2)^2 + 5(-1/2) - 1 = -9/4, so vertex \left( -1/2, -9/4 \right)..

Q: How do I complete the square for this equation?

Divide by 5: x^2 + x - 1/5 = 0. Then (x + 1/2)^2 - 1/4 - 1/5 = 0 \rightarrow (x + 1/2)^2 = 9/20..

Q: How can I sketch the graph quickly?

It's an upward-opening parabola (a = 5 > 0), vertex at -1/2, -9/4 , and x-intercepts at x = (-5 \pm 3\sqrt{5})/10 ; plot vertex and intercepts for shape.

Answer

Subtract 1 from both sides: \[5x^{2}+5x-1=0\]

Detailed Explanation

Write the given equation exactly as provided:
\[5x^2 + 5x = 1\]
Standard form for a quadratic is ax^2 + bx + c = 0, so we need zero on the right-hand side. To achieve this, subtract 1 from both sides of the equation (perform the same operation on each side to preserve equality):
\[5x^2 + 5x – 1 = 0\]

Explanation: subtracting 1 from the right-hand side (1) yields 0, and subtracting 1 from the left-hand side changes the constant term from 0 to -1, giving the left-hand expression a, b, and c terms explicitly.
Optional normalization: if you prefer the leading coefficient to be 1, divide every term by 5. This produces an equivalent equation with a = 1:
\[x^2 + x – \tfrac{1}{5} = 0\]

However, the standard form with integer coefficients is:

\[5x^2 + 5x – 1 = 0\]

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Algebra FAQs

How do you write the equation in standard form?

Move all terms to one side so the equation equals zero: \(5x^2 + 5x - 1 = 0.\)

How do I know this is a quadratic?

A quadratic has the form \(ax^2 + bx + c\) with \(a \ne 0\). Here \(a = 5\), \(b = 5\), \(c = -1\), so it is quadratic.

Can the equation be simplified by factoring out a common factor?.

No integer common factor > 1 exists (gcd\(5,5,-1\)=1). Dividing by 5 gives fractions: \(x^2 + x - 1/5 = 0\)..

How do I find the roots using the quadratic formula?

Use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=5,b=5,c=-1\). Discriminant \(45\), so \(x = \frac{-5 \pm \sqrt{45}}{10} = \frac{-5 \pm 3\sqrt{5}}{10}\)..

How many real solutions does the equation have?

Discriminant \( \Delta = 45 > 0 \), so there are two distinct real roots.

Can the quadratic be factored over the integers or rationals?

What is the vertex and axis of symmetry?

Axis: \(x = -b/(2a) = -5/10 = -1/2\). Vertex: plug in to get \(y = 5(-1/2)^2 + 5(-1/2) - 1 = -9/4\), so vertex \( \left( -1/2, -9/4 \right)\)..

How do I complete the square for this equation?

Divide by 5: \(x^2 + x - 1/5 = 0\). Then \((x + 1/2)^2 - 1/4 - 1/5 = 0 \rightarrow (x + 1/2)^2 = 9/20\)..

How can I sketch the graph quickly?

It's an upward-opening parabola (a = 5 > 0), vertex at \( -1/2, -9/4 \), and x-intercepts at \( x = (-5 \pm 3\sqrt{5})/10 \); plot vertex and intercepts for shape.

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