Analyzing the Quadratic Equation for Integral Roots

We are given the quadratic equation: \(x^2 + 7x - 14(k^2 - 7/8) = 0\), where \(k\) is an integer. We need to determine if this equation has any integral roots for integer values of \(k\).

A quadratic equation of the form \(ax^2 + bx + c = 0\) has integral roots if and only if:

1. The discriminant, \(D = b^2 - 4ac\), is a perfect square.
2. The values obtained from the quadratic formula, \(x = \frac{-b \pm \sqrt{D}}{2a}\), are integers.

Let's identify the coefficients \(a\), \(b\), and \(c\) from the given equation:

• \(a = 1\)
• \(b = 7\)
• \(c = -14(k^2 - 7/8) = -14k^2 + 14 \times \frac{7}{8} = -14k^2 + \frac{98}{8} = -14k^2 + \frac{49}{4}\)

So the equation is \(x^2 + 7x + (-14k^2 + 49/4) = 0\).

Calculating the Discriminant

Now, let's calculate the discriminant \(D\):

\(D = b^2 - 4ac\)

\(D = (7)^2 - 4(1)(-14k^2 + 49/4)\)

\(D = 49 - 4(-14k^2) - 4(49/4)\)

\(D = 49 + 56k^2 - 49\)

\(D = 56k^2\)

When is the Discriminant a Perfect Square?

For the roots to be rational (a necessary condition for integral roots), the discriminant \(D\) must be a perfect square. We have \(D = 56k^2\). We need to check for which integer values of \(k\) this expression is a perfect square.

Let's look at the prime factorization of 56: \(56 = 8 \times 7 = 2^3 \times 7^1\).

So, \(D = 2^3 \times 7^1 \times k^2\).

For a number to be a perfect square, the exponents of all prime factors in its prime factorization must be even.

Let the prime factorization of the integer \(k\) be \(k = \pm 2^{a_2} 7^{a_7} m\), where \(m\) is an integer not divisible by 2 or 7, and \(a_2, a_7 \ge 0\).

Then \(k^2 = 2^{2a_2} 7^{2a_7} m^2\). The exponents \(2a_2\), \(2a_7\), and the exponents of prime factors in \(m^2\) are all even.

Now substitute this into the discriminant:

\(D = (2^3 \times 7^1) \times (2^{2a_2} 7^{2a_7} m^2)\)

\(D = 2^{3+2a_2} \times 7^{1+2a_7} \times m^2\)

For \(D\) to be a perfect square, the exponents of all prime factors must be even.

• The exponent of 2 is \(3 + 2a_2\). Since \(a_2\) is an integer, \(2a_2\) is an even number. An odd number (3) plus an even number (\(2a_2\)) is always an odd number. So, \(3 + 2a_2\) is always odd for any integer \(a_2 \ge 0\).
• The exponent of 7 is \(1 + 2a_7\). Since \(a_7\) is an integer, \(2a_7\) is an even number. An odd number (1) plus an even number (\(2a_7\)) is always an odd number. So, \(1 + 2a_7\) is always odd for any integer \(a_7 \ge 0\).

This shows that for any integer \(k\) where \(k \neq 0\), the discriminant \(D = 56k^2\) will have odd exponents for its prime factors 2 and 7 (at least), and thus cannot be a perfect square.

The only case where \(D=56k^2\) can be a perfect square is if \(D=0\). This happens when \(56k^2=0\), which implies \(k=0\) (since 56 is non-zero).

Roots when Discriminant is a Perfect Square (Only k=0)

If \(k=0\), the equation becomes \(x^2 + 7x - 14(0^2 - 7/8) = 0\).

\(x^2 + 7x - 14(-7/8) = 0\)

\(x^2 + 7x + \frac{98}{8} = 0\)

\(x^2 + 7x + \frac{49}{4} = 0\)

Multiplying the entire equation by 4 to remove the fraction:

\(4(x^2 + 7x + 49/4) = 4(0)\)

\(4x^2 + 28x + 49 = 0\)

This is a perfect square trinomial: \((2x)^2 + 2(2x)(7) + 7^2 = 0\), which is \((2x + 7)^2 = 0\).

The roots are given by \(2x + 7 = 0\), so \(2x = -7\), and \(x = -7/2\).

The roots are \(-7/2\) (a repeated root). Since \(-7/2 = -3.5\), the roots are rational but not integers.

Roots when Discriminant is Not a Perfect Square (For integer k ≠ 0)

As shown, for any integer \(k \neq 0\), the discriminant \(D = 56k^2\) is not a perfect square. This means that \(\sqrt{D} = \sqrt{56k^2} = |k|\sqrt{56} = |k| \sqrt{4 \times 14} = |k| 2\sqrt{14}\). Since 14 is not a perfect square, \(\sqrt{14}\) is an irrational number.

Therefore, for \(k \neq 0\), \(\sqrt{D}\) is an irrational number.

The roots are given by the quadratic formula: \(x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{56k^2}}{2(1)} = \frac{-7 \pm |k|2\sqrt{14}}{2} = \frac{-7}{2} \pm |k|\sqrt{14}\).

Since \(|k|\sqrt{14}\) is irrational for any integer \(k \neq 0\), the roots \(x = -7/2 \pm |k|\sqrt{14}\) are irrational.

Irrational numbers cannot be integral.

Conclusion on Integral Roots

We have examined all possible integer values for \(k\):

• If \(k=0\), the roots are \(-7/2\), which are not integral.
• If \(k \neq 0\) (for any other integer \(k\)), the roots are irrational, and therefore not integral.

In every case where \(k\) is an integer, the roots of the equation \(x^2 + 7x - 14(k^2 - 7/8) = 0\) are not integers.

Therefore, the equation has no integral roots.

Revision Table: Quadratic Equation Roots

Additional Information: Properties of Integral Roots

For a quadratic equation \(ax^2 + bx + c = 0\) with integer coefficients \(a, b, c\), if the roots are integral, say \(r_1\) and \(r_2\), then:

• The sum of roots: \(r_1 + r_2 = -b/a\). This must be an integer since \(a\) and \(b\) are integers and the sum is an integer.
• The product of roots: \(r_1 \times r_2 = c/a\). This must be an integer since \(a\) and \(c\) are integers and the product is an integer.

In our equation, \(a=1\) and \(b=7\). The sum of roots is \(-b/a = -7/1 = -7\), which is an integer. This condition alone is not sufficient to guarantee integral roots, as rational roots can also sum to an integer (e.g., \(-3.5 + -3.5 = -7\), as seen when \(k=0\)).

The product of roots is \(c/a = (-14k^2 + 49/4)/1 = -14k^2 + 49/4\). For integral roots, this product must be an integer. However, \(-14k^2 + 49/4\) is an integer only if \(49/4\) is an integer, which it is not, or if \(-14k^2\) has a fractional part that cancels out \(49/4\). Let's check when \(-14k^2 + 49/4\) is an integer for integer \(k\).

\(-14k^2\) is always an integer for integer \(k\). For the sum \(-14k^2 + 49/4\) to be an integer, \(49/4\) must be an integer. It is not (\(49/4 = 12.25\)). So, \(-14k^2 + 49/4\) is never an integer for any integer \(k\).

Since the product of integral roots must be an integer, and \(-14k^2 + 49/4\) is never an integer for integer \(k\), this also confirms that the equation can have no integral roots for any integer \(k\).

This alternative approach using the product of roots property strongly supports the conclusion derived from analyzing the discriminant.