QUADRATIC EQUATIONS-Exercise 4.3
Maths - Exercise
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
- \(2x^2 – 3x + 5 = 0\)
Solution:
Given: \(2x^{2} - 3x + 5 = 0\)
To determine the nature of the roots, calculate the discriminant \[D = b^2 - 4ac\] where \(a = 2\), \(b = -3\), and \(c = 5\).
Calculate \[ \begin{aligned} D &= (-3)^2 - 4 \times 2 \times 5 \\ &= 9 - 40 \\ &= -31 \end{aligned} \]
Since \(D < 0\), the quadratic equation has no real roots. Instead, it has two complex conjugate roots.
Therefore, no real roots exist for this equation.
- \(3x^2 – 4 \sqrt{3} x + 4 = 0\)
Solution:
Given: \(3x^{2} - 4\sqrt{3} x + 4 = 0\)
Calculate the discriminant \[D = b^{2} - 4ac\] with \(a = 3\), \(b = -4\sqrt{3}\), and \(c = 4\).
Calculate \[ \begin{aligned} D &= (-4\sqrt{3})^2 - 4 \times 3 \times 4 \\ &= 48 - 48 \\ &= 0 \end{aligned} \]
Since \(D = 0\), the quadratic has equal real roots. The roots are given by \[ \begin{aligned} x &=\dfrac{-b}{2a} \\ &= \dfrac{-(-4\sqrt{3})}{2 \times 3} \\ &= \dfrac{4\sqrt{3}}{6} \\ &= \dfrac{2}{\sqrt{3}} \end{aligned} \]
Therefore, the equation has two equal real roots, both equal to \(\dfrac{2}{\sqrt{3}}\).
- \(
2x^2 – 6x + 3 = 0\)
Solution:
Given: \(2x^{2} - 6x + 3 = 0\)
Calculate the discriminant \[D = b^{2} - 4ac\] where \(a = 2\), \(b = -6\), and \(c = 3\)
Compute \[ \begin{aligned} D &= (-6)^{2} - 4 \times 2 \times 3 \\ &= 36 - 24 \\ &= 12 \end{aligned} \]
Since \(D > 0\), the quadratic equation has two distinct real roots.
The roots are given by \[ \begin{aligned} x &= \frac{-b \pm \sqrt{D}}{2a} \\ &= \frac{-(-6) \pm \sqrt{12}}{2 \times 2} \\ &= \frac{6 \pm 2\sqrt{3}}{4}\\ &= \frac{3 \pm \sqrt{3}}{2}. \end{aligned} \]
Thus, the roots are \(\frac{3 + \sqrt{3}}{2}\) and \(\frac{3 - \sqrt{3}}{2}\).
Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
- \( 2x^2 + kx + 3 = 0\)
Solution:
Given: \(2x^2 + kx + 3 = 0\)
For the quadratic equation to have two equal real roots, the discriminant must be zero: \[D = b^2 - 4ac = 0\]
Here, \(a = 2\), \(b = k\), and \(c = 3\).
Set \(D = 0\): \[k^2 - 4 \times 2 \times 3 = 0\]Simplify: \[ \begin{aligned} k^2 - 24 &= 0\\ k^2 &= 24\\ k &= \pm \sqrt{24} \\ &= \pm 2\sqrt{6} \end{aligned} \]
Thus, the values of \(k\) are \(2\sqrt{6}\) and \(-2\sqrt{6}\).
- \(kx (x – 2) + 6 = 0\)
Solution:
Given: \(kx(x - 2) + 6 = 0\)
First expand the equation: \[kx^2 - 2kx + 6 = 0\] This is a quadratic equation with \(a = k\), \(b = -2k\), and \(c = 6\).
For equal roots, the discriminant must be zero: \[D = b^2 - 4ac = 0\] Substitute the coefficients: \[ \begin{aligned} (-2k)^2 - 4(k)(6) &= 0\\ 4k^2 - 24k &= 0\\ 4k(k - 6) = &0\\ \end{aligned} \]
Thus, \[ 4k = 0\] or \[ \begin{aligned} k - 6 &= 0\\ k&=6 \end{aligned} \]
Solutions: \(k = 0\) or \(k = 6\).
Note that \(k = 0\) makes the equation linear (\(6 = 0\)), which has no solution, so discard it.Therefore, the only valid value is \(k = 6\).
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Solution:
Given: Rectangular mango grove with length twice its breadth and area 800 m². Find dimensions if possible.
Let breadth be \(x\) m.
Then length is \(2x\) m.
Area equation: \(x \times 2x = 800\).
Simplify: \[ \begin{aligned} 2x^2 &= 800\\ x^2 &= 400\\ x &= \pm \sqrt{400} \\ &= \pm 20 \end{aligned} \]
Since dimensions must be positive, discard \(x = -20\).
Thus, breadth \(x = 20\) m and
length
\(2x = 40\) m.
Yes, it is possible.
Dimensions:
length 40 m, breadth 20 m.
Verification: \(40 \times 20
= 800\) m².
Q.4 Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Given: Sum of ages of two friends is 20 years. Four years ago, product of their ages was 48. Determine if possible and find present ages.
Let present age of first friend be \(x\) years.
Then second friend's age is \(20 - x\)
years.
Four years ago, their ages were \(x - 4\) and \(16 - x\) years.
Product four years ago: \((x - 4)(16 - x) = 48\). \[ \begin{aligned} (x - 4)(16 - x) &= 48\\ 16x - x^2 - 64 + 4x &= 48\\ -x^2 + 20x - 112 &= 0\\ x^2 - 20x + 112 &= 0 \end{aligned} \]
Discriminant: \[ \begin{aligned} D &= b^2 - 4ac \\ &= (-20)^2 - 4(1)(112) \\ &= 400 - 448 \\ &= -48 \end{aligned} \]
Since \(D < 0\), no real solutions exist. Ages cannot be negative or imaginary.
Therefore, the given situation is not possible.
Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution:
Given: Rectangular park with perimeter 80 m and area 400 m². Find dimensions if possible.
Let breadth be \(x\) m.
Then length is \(40 - x\) m,
since perimeter \(2(l + b) = 80\) gives \(l + b = 40\).
Area equation:
\[
\begin{aligned}
x(40 - x) &= 400\\
40x - x^2 &= 400\\
x^2 - 40x + 400 &= 0
\end{aligned}
\]
Discriminant: \[ \begin{aligned} D &= (-40)^2 - 4(1)(400)\\ &= 1600 - 1600 \\ &= 0 \end{aligned} \]
Since \(D = 0\), equal real roots exist.
Root:
\[
\begin{aligned}
x &= \dfrac{-b}{2a} \\
&= \dfrac{40}{2} \\
&= 20
\end{aligned}
\]
Thus, breadth \(x = 20\) m and length \(40 - 20 = 20\) m.
Yes, it is possible.
The park is a square with sides 20 m.
Verification: Perimeter
\(80\) m, area \(400\) m².
