What is the nature of roots of the quadratic equation if the value of its discriminant is greater than zero and a perfect square?

\(x^2 + 3x = -2\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[x^2 + 3x + 2 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = 3; \qquad c = 2\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (3)^2 - 4(1)(2) \\ &= 9 - 8 \\ &= 1 \end{align*}

We know that \(1 > 0\) and is a perfect square.

We have calculated that \(Δ > 0\) and is a perfect square, therefore we can conclude that the roots are real, unequal and rational.

\(x^2 + 9 = 6x\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[x^2 - 6x + 9 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = -6; \qquad c = 9\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (-6)^2 - 4(1)(9) \\ &= 36 - 36 \\ &= 0 \end{align*}

We have calculated that \(Δ = 0\) therefore we can conclude that the roots are real and equal.

\(6y^2 - 6y - 1 = 0\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[6y^2 - 6y - 1 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 6; \qquad b = -6; \qquad c = -1\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (-6)^2 - 4(6)(-1) \\ &= 36 + 36 \\ &= 72 \end{align*}

We know that \(72 > 0\) and is not a perfect square.

We have calculated that \(Δ > 0\) and is not a perfect square, therefore we can conclude that the roots are real, unequal and irrational.

\(4t^2 - 19t - 5 = 0\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[4t^2 - 19t - 5 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 4; \qquad b = -19; \qquad c = -5\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (-19)^2 - 4(4)(-5) \\ &= 361 + 80 \\ &= 441 \end{align*}

We know that \(441 > 0\) and is a perfect square.

We have calculated that \(Δ > 0\) and is a perfect square, therefore we can conclude that the roots are real, unequal and rational.

\(z^2 = 3\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[z^2 - 3 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = 0; \qquad c = -3\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (0)^2 - 4(1)(-3) \\ &= 0 + 12 \\ &= 12 \end{align*}

We know that \(12 > 0\) and is not a perfect square.

We have calculated that \(Δ > 0\) and is not a perfect square, therefore we can conclude that the roots are real, unequal and irrational.

\(0 = p^2 + 5p + 8\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[p^2 + 5p + 8 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = 5; \qquad c = 8\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (5)^2 - 4(1)(8) \\ &= 25 - 32 \\ &= -7 \end{align*}

We know that \(-7 < 0\).

We have calculated that \(Δ < 0\), therefore we can conclude that the roots are non-real.

\(x^2 = 36\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[x^2 - 36 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = 0; \qquad c = -36\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (0)^2 - 4(1)(-36) \\ &= 0 + 144 \\ &= 144 \end{align*}

We know that \(144 > 0\) and is a perfect square.

We have calculated that \(Δ > 0\) and is a perfect square, therefore we can conclude that the roots are real, unequal and rational.

\(4m + m^2 = 1\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[m^2 + 4m - 1 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = 4; \qquad c = -1\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (4)^2 - 4(1)(-1) \\ &= 16 + 4 \\ &= 20 \end{align*}

We know that \(20 > 0\) and is not a perfect square.

We have calculated that \(Δ > 0\) and is not a perfect square, therefore we can conclude that the roots are real, unequal and irrational.

\(11 - 3x + x^2 = 0\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[x^2 - 3x + 11 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 1; \qquad b = -3; \qquad c = 11\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (-3)^2 - 4(1)(11) \\ &= 9 - 44 \\ &= -35 \end{align*}

We know that \(-35 < 0\).

We have calculated that \(Δ < 0\), therefore we can conclude that the roots are non-real.

\(y^2 + \dfrac{1}{4} = y\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[4y^2 - 4y + 1 = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = 4; \qquad b = -4; \qquad c = 1\]

Write down the formula and substitute values.

\begin{align*} Δ &= b^2-4ac \\ &= (-4)^2 - 4(4)(1) \\ &= 16 - 16 \\ &= 0 \end{align*}

We have calculated that \(Δ = 0\), therefore we can conclude that the roots are real and equal.

Show that the discriminant is given by: \(\Delta ={k}^{2}+6bk+{b}^{2}+8\)

We write the equation in standard form \(ax^2+bx+c=0\):

\[(k + 1)x^2 + (b + 3k)x - 2 + 2k = 0\]

Identify the coefficients to substitute into the formula for the discriminant

\[a = k + 1; \qquad b = b + 3k; \qquad c = -2 + 2k\]

Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (b + 3k)^2 - 4(k + 1)(2k - 2) \\ &= b^{2} + 6bk + 9k^{2} -4(2k^{2} - 2) \\ &= b^{2} + 6bk + 9k^{2} -8k^{2} + 8 \\ &= k^{2} + 6bk + b^{2} + 8 \end{align*}

If \(b=0\), discuss the nature of the roots of the equation.

When \(b=0\) the discriminant is:

\[Δ = k^{2} + 8\]

This is positive for all values of \(k\) and greater than \(\text{0}\) for all values of \(k\).

For example if \(k = 0\) then \(Δ = 8\), if \(k = -1\) then \(Δ = 9\) and if \(k = 1\) then \(Δ = 9\).

So the roots are real and unequal. We cannot say if the roots are rational or irrational since this depends on the exact value of \(k\).

If \(b=2\), find the value(s) of \(k\) for which the roots are equal.

When \(b=2\) the discriminant is:

\begin{align*} Δ & = k^{2} + 6(2)k + (2)^{2} + 8 \\ & = k^{2} + 12k + 12 \end{align*}

We set this equal to \(\text{0}\) since we want to find the values of \(k\) that will make the roots equal.

\begin{align*} 0 & = k^{2} + 12k + 12 \\ k & = \dfrac{-12 \pm \sqrt{(12)^{2} - 4(1)(12)}}{2(1)} \\ & = \dfrac{-12 \pm \sqrt{144 - 48}}{2} \\ & = \dfrac{-12 \pm \sqrt{96}}{2} \\ & = \dfrac{-12 \pm 4\sqrt{6}}{2} \\ k = -6 + 2\sqrt{6} & \text{ or } k = -6 - 2\sqrt{6} \end{align*}

The roots will be equal if \(k = -6 \pm 2\sqrt{6}\).

Show that \({k}^{2}{x}^{2}+2=kx-{x}^{2}\) has non-real roots for all real values for \(k\).

[IEB, Nov. 2002, HG]

\begin{align*} k^2 x^2 + x^2 - kx +2 &= 0\\ a &= (k^2+1)\\ b&= -k\\ c&=2\\ \Delta &= b^2-4ac\\ &= (-k)^2 - 4(k^2+1)(2)\\ &=k^2-8k^2-8\\ &=-7k^2 - 8\\ &=-(7k^2 + 8)\\ \Delta & < 0 \end{align*}Therefore the roots are non-real.

Find the greatest value of value \(k\) such that \(k \in \mathbb{Z}\).

\begin{align*} x^2 +12x &= 3kx^2 +2\\ 3kx^2 -x^2 - 12x+2&=0\\ x^2(3k-1)-12x+2&=0\\ a&=3k-1\\ b&=-12\\ c&=2\\ \text{Given } \Delta &\geq 0\\ \therefore b^2-4ac &\geq 0\\ \therefore (-12)^2 - 4(3k-1)(2) &\geq 0\\ 144 -24k + 8 &\geq 0 \\ 152 - 24k &\geq 0 \\ 152 &\geq 24k \\ \frac{19}{3} &\geq k \\ \therefore k &\leq \frac{19}{3} \\ \therefore k &\leq 6\frac{1}{3} \end{align*}

Since \(k\) needs to be an integer, the greatest value of \(k\) is \(\text{6}\).

Find one rational value of \(k\) for which the above equation has rational roots.

For rational roots we need \(\Delta\) to be a perfect square.

\begin{align*} \Delta &= b^2-4ac \\ &=152-24k\\ \text{if } k&=\frac{1}{3}\\ \Delta &= 152 - 24\left(\frac{1}{3}\right)\\ &=152-8\\ &=144\\ &=12^2 \end{align*}

This is a perfect square. Therefore if \(k=\frac{1}{3}\), the roots will be rational.

Find a value of \(k\) for which the roots are equal.

We first need to write the equation in standard form:

\begin{align*} k(2x - 5) & = x^2-4 \\ 2kx - 5k & = x^2-4 \\ 0 & = x^{2} - 4 - 2kx + 5k \\ 0 & = x^{2} - 2kx + 5k - 4 \end{align*}

Next we note that \(a = 1; \qquad b = -2k; \qquad c = 5k - 4\).

Now we can find the discriminant:

\begin{align*} Δ &= b^2-4ac \\ &= (-2k)^2 - 4(1)(5k - 4) \\ &= 4k^{2} - 20k + 16 \end{align*}

To find the values of \(k\) that make the roots equal, we set this equal to \(\text{0}\) and solve for \(k\):

\begin{align*} 0 &= 4k^{2} - 20k + 16 \\ &= k^{2} - 5k + 4 \\ &= (k - 4)(k - 1) \\ k = 4 & \text{ or } k = 1 \end{align*}

Find an integer \(k\) for which the roots of the equation will be rational and unequal.

The discriminant is:

\[\Delta = 4k^{2} - 20k + 16\]

To find a value of \(k\) that makes the roots rational and unequal the discriminant must be greater than \(\text{0}\) and a perfect square.

We try setting the discriminant equal to \(\text{1}\):

\begin{align*} \Delta & = 4k^{2} - 20k + 16 \\ 1 & = 4k^{2} - 20k + 16 \\ 0 & = 4k^{2} - 20k + 15 \end{align*}

This does not give an integer value of \(k\) so we try \(\text{4}\):

\begin{align*} 4 & = 4k^{2} - 20k + 16 \\ 0 & = 4k^{2} - 20k + 12 \\ & = k^{2} - 5k + 3 \end{align*}

This does not give an integer value of \(k\) so we try \(\text{9}\):

\begin{align*} 9 & = 4k^{2} - 20k + 16 \\ 0 & = 4k^{2} - 20k + 5 \end{align*}

This does not give an integer value of \(k\) so we try \(\text{16}\):

\begin{align*} 16 & = 4k^{2} - 20k + 16 \\ 0 & = 4k^{2} - 20k \\ & = k^{2} - 5k \\ k = 0 & \text{ or } k = 5 \end{align*}

So if \(k=0\) or \(k=5\) the roots will be rational and unequal.

Prove that the roots of the equation \({x}^{2}-\left(a+b\right)x+ab-{p}^{2}=0\) are real for all real values of \(a\), \(b\) and \(p\).

We need to prove that \(\Delta \geq 0\).

\begin{align*} \Delta &=(-a-b)^2 - 4(ab-p^2)\\ &=a^2 + 2ab + b^2 - 4ab + 4p^2\\ &=a^2 - 2ab + b^2 + 4p^2\\ &=(a-b)^2 + 4p^2 \end{align*}

\(\Delta \geq 0\) for all real values of \(a\), \(b\) and \(p\). Therefore the roots are real for all real values of \(a\), \(b\) and \(p\).

When will the roots of the equation be equal?

The roots are equal when \(\Delta = 0\), that is when \(a=b\) and \(p=0\).

If \(b\) and \(c\) can take on only the values \(\text{1}\), \(\text{2}\) or \(\text{3}\), determine all pairs (\(b; c\)) such that \({x}^{2}+bx+c=0\) has real roots.

[IEB, Nov. 2005, HG]

We need to find the values of \(a\), \(b\) and \(c\) for which \(\Delta \geq 0\).

\begin{align*} a&=1\\ b&=\text{1,2} \text{ or }3\\ c&=\text{1,2} \text{ or }3\\ \\ \Delta &= b^2-4ac\\ &= b^2 - 4(1)c \end{align*}

Possible pair values of \((b;c)\): \((1;1),~(1;2),~(1;3),~(2;1),~(2;2),~(2;3),~(3;1),~(3;2),~(3;3)\). Corresponding values of \(\Delta\): \((\Delta < 0),~(\Delta < 0),~(\Delta < 0),~(\Delta = 0),~(\Delta < 0),~(\Delta < 0),~(\Delta >0),~(\Delta > 0),~(\Delta < 0)\)

\(\Delta ≥ 0\) (and therefore the roots are real) for \((b;c) = (2;1),~(3;1),~(3;2)\)