We know that a quadratic equation is a second degree polynomial equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, x is the unknown variable and a ≠ 0. For the equation ax2 + bx + c = 0, the discriminant is given by D = b2 – 4ac. It is also denoted by ∆. A quadratic equation has 2 roots. It will be real or imaginary. In this article we discuss the nature of roots depending upon coefficients and discriminant. Show If α and β are the values of x which satisfy the quadratic equation, α and β are called the roots of the quadratic equation. Roots are given by the equation (-b±√(b2-4ac))/2a. The nature of the roots depends on the discriminant. Nature of Roots depending upon DiscriminantAccording to the value of discriminant, we shall discuss the following cases about the nature of roots. Case 1: D = 0 If the discriminant is equal to zero (b2 – 4ac = 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are real and equal. In this case, the roots are x = -b/2a. The graph of the equation touches the X axis at a single point. Case 2: D > 0 If the discriminant is greater than zero (b2 – 4ac > 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are real and unequal. The graph of the equation touches the X-axis at two different points. Case 3: D < 0 If the discriminant is less than zero (b2 – 4ac < 0), a, b, c are real numbers, a≠0, then the roots of the quadratic equation ax2 + bx + c = 0, are imaginary and unequal. The roots exist in conjugate pairs. The graph of the equation does not touch the X-axis. Case 4: D > 0 and perfect square If D > 0 and a perfect square, then the roots of the quadratic equation are real, unequal and rational. Case 5: D > 0 and not a perfect square If D > 0 and not a perfect square, then the roots of the quadratic equation are real, unequal and irrational. We can summarize all the above cases in the table below.
Nature of Roots depending upon coefficientsDepending upon the nature of the coefficients of the quadratic equation, we can summarize the following.
Bridge Course – Nature of Roots of Quadratic Equations Also Read Quadratic inequalities Solved ExamplesExample 1: The roots of the quadratic equation 3x2-10x+3 = 0 are a) real and equal b) imaginary c) real, unequal and rational d) none of these Solution: Given equation 3x2-10x+3 = 0 Here discriminant, D = b2-4ac => (-10)2 – 4×3×3 = 100 – 36 = 64 D is positive and a perfect square. So the roots of the quadratic equation are real, unequal and rational. Hence option c is the answer. Example 2: Find the value of p if the equation 3x2-18x+p = 0 has real and equal roots. a) 27 b) 18 c) 9 d) none of these Solution: Given 3x2-18x+p = 0 has real and equal roots. => b2-4ac = 0 =>(-18)2-4×3×p = 0 => 324 – 12p = 0 => p = 324/12 = 27 Hence option a is the answer. Example 3: The quadratic equation with real coefficients when one of its root is (3+2i) is Solution: Given one root is 3+2i. Complex roots occur in conjugate pairs. So other root = 3-2i Sum of roots = 6 Product of roots = (3+2i)(3-2i) = 13 Required equation is x2-(Sum)x+Product = 0 => x2-6x+13 = 0 Example 4: Show that the equation 3x2+4x+6 = 0 has no real roots. Solution: Given equation 3x2+4x+6 = 0 Here a = 3, b = 4, c = 6 Discriminant D = b2-4ac => 42-4×3×6 = 16-72 = -56 Since D<0, the roots are imaginary. Hence the equation has no real roots. Video Lesson – Nature of Roots
The discriminant of a quadratic equation is given by D = b2 – 4ac.
If discriminant, D = 0, then the roots are real and equal.
If discriminant, D>0, then the roots are real and unequal.
If discriminant, D<0, then the roots are imaginary and unequal.
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How to Determine Nature of Roots -- Quadratic Equations
To determine the nature of roots of quadratic equations (in the form ax^2 + bx +c=0) , we need to caclulate the discriminant, which is b^2 - 4 a c. When discriminant is greater than zero, the roots are unequal and real. When discriminant is equal to zero, the roots are equal and real. When discriminant is less than zero, the roots are imaginary.
\(x^2 + 3x = -2\)
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 \((b;c) = (2;1),~(3;1),~(3;2)\) |
Which of the following best describe the discriminant and nature of the roots of the equation
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