Grade 10 – Math – Quadratic Equation- Important Questions and Answers

    Quadratic equation

    Question 1: Three consecutive positive integers are such that the sum of the product of the first square and the other two is 7. Determine this position as a quadratic equation.

    Solution : Suppose three consecutive positive integers are $ x, x + 1 $ and $ x + 2 $. Then, according to the question,
    \[\begin{array}{l} x^{2}+(x+1)(x+2)=7 \\\Rightarrow x^{2}+x(x+2)+1(x+2)=7 \\\Rightarrow x^{2}+x^{2}+2 x+x+2=7 \\\Rightarrow 2 x^{2}+3 x-5=0\end{array}\]
    Which is the intended quadratic equation.

    Question 2: Denominator of a fraction is 1 more than two times its numerator. The sum of the fraction and its inverse is $ 2 \ frac {16} {21} $. Represent this condition as a quadratic equation.

    Solution: Let the numerator of the fraction be $ x $. Then, according to the question, denominator $ = 2 x + 1 $
    \[ \therefore \text {fraction} = \frac {x} {2 x + 1} \]
    \[\text{Its inverse} = \frac {2 x + 1} {x} \]
    Again, according to the question, \[ \frac {x} {2 x + 1} + \frac {2 x + 1} {x} = \frac {58} {21} \quad \left[\because 2 \frac{16}{21}=\frac{58}{21} \right] \]
    \[\begin{aligned}
    &\Rightarrow \quad \frac{x^{2}+(2 x+1)^{2}}{(2 x+1) x}=\frac{58}{21}\\
    &\Rightarrow \quad{x^{2}+4 x^{2}+4 x+1}{2 x^{2}+x}=\frac{58}{21}\\
    &\Rightarrow \quad \frac{5 x^{2}+4 x+1}{2 x^{2}+x}=\frac{58}{21}\\
    &\Rightarrow \quad 58\left(2 x^{2}+x\right)=21\left(5 x^{2}+4 x+1\right)\\
    &\Rightarrow \quad 116 x^{2}+58 x=105 x^{2}+84 x+21\\
    &\Rightarrow \quad 11 x^{2}-26 x-21=0\\
    \end{aligned}\]
    Which is the intended quadratic equation.

    Question 3. Akash wants to fit 3 rods together into a right angled triangle. The hypotenuse is 2 cm longer than the base and 1 cm longer than the apex. Represent this condition as a quadratic equation.

    Solution: Let the length of the hypotenuse be $ x $ cm.
    base = (x-2) cm
    apex = (x-1) cm
    From the Pythagoras theorem, \[ \begin {aligned} &(\text {base}) ^ {2} + (\text {apex}) ^ {2} = (\text {hypotenuse}) ^ {2} \\ &\Rightarrow (x-2) ^ {2} + (x-1) ^ {2} = x ^ {2} \\ &\Rightarrow x ^ {2} -4 x + 4 + x ^ {2} -2 x + 1 = x ^ {2} \\ &\Rightarrow x ^ {2} -6 x + 5 = 0 \end {aligned} \]
    Which is the exact quadratic equation.

    Question 4: x(x + 1) + 5 = (x + 2) (x – 2) is a quadratic equation. True/False?

    Answer: False

    Question 5: $(x + 2) ^3 = x ^3 + 4$ is a ………….. equation.

    Answer: Quadratic.

    Solution of Quadratic Equation by Factorisation

    Question 1: Three consecutive natural numbers are such that the square of the middle number exceeds the difference
    of the squares of the other two by $60 .$ Find the numbers .
    Solution:
    Let three consecutive natural numbers are $x-1, x$ and $x+1$ According to question,
    \[
    \begin{aligned}
    &x^{2}-\left[(x+1)^{2}-(x-1)^{2}\right] =60 \\
    &\Rightarrow x^{2}-[(x+1-x+1)(x+1+x-1)] =60 \\
    &\Rightarrow x^{2}-4 x-60 =0\\
    &\Rightarrow x^{2}-10 x+6 x-60=0 \\
    &\Rightarrow x(x-10)+6(x-10)=0 \\
    &\Rightarrow (x+6)(x-10)=0 \\
    &\Rightarrow x=10 \quad (x=-6 \text{ rejected} )
    \end{aligned}
    \]
    Hence, the numbers are 9,10 and $11 .$

    Question 2: A shopkeeper buys some books for $80 .$ If he had bought 4 more books for the same amount, each book
    would have cost ₹1 less. Find the number of books he bought. $(2012 \mathrm{D})$

    Solution:
    Let the number of books he bought $=x$
    Increased number of books he had bought $=x+4$
    Total amount $=₹ 80$
    According to the problem,
    \begin{aligned}
    &\frac{80}{x}-\frac{80}{x+4}=1\\
    &\Rightarrow \frac{80(x+4-x)}{x(x+4)}=1\\
    &\Rightarrow x(x+4)=320\\
    &\Rightarrow x^{2}+4 x-320=0\\
    &\Rightarrow x^{2}+20 x-16 x-320=0\\
    &\Rightarrow x(x+20)-16(x+20)=0\\
    &\Rightarrow(x+20)(x-16)=0\\
    &\Rightarrow x+20=0 \text{ or } x-16=0\\
    &\Rightarrow x=-20 \ldots \text{(neglected) or } x=16\\
    \end{aligned}
    $\therefore \text{Number of books he bought} =16$

    Question 3: A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for treatment. Due to this, the plane got delayed by half an hour. To reach the destination 1500 km away in time, so that the passengers could catch the connecting flight, the speed of the plane was increased by 250 km/hour than the usual speed. Find the usual speed of the plane. What value is depicted in this question?

    Solution:

    Let the usual speed of plane be $x \mathrm{ km} / \mathrm{hr}$. Time taken to cover $1500 \mathrm{ km}$ with usual speed $=\frac{1500}{x}$ hours.
    When the speed of plane is increased, then
    \[\text{new speed} =(x+250) \mathrm{ km} / \mathrm{hr} \]
    Time taken to cover $1500 \mathrm{ km}$ with the new speed
    \[(x+250) \mathrm{ km} / \mathrm{hr}=\frac{1500}{x+250}\]
    According to question,
    \[
    \begin{aligned}
    &\frac{1500}{x}=\frac{1500}{x+250}+\frac{1}{2} \\
    &\Rightarrow \frac{1500}{x}-\frac{1500}{x+250}=\frac{1}{2}\\
    &\Rightarrow \frac{1500 x+1500 \times 250-1500 x}{x(x+250)}=\frac{1}{2} \\
    &\Rightarrow \frac{1500 \times 250}{x^{2}+250 x}=\frac{1}{2}\\
    &\Rightarrow x^{2}+250 x=750000 \\
    &\Rightarrow x^{2}+250 x-750000=0\\
    &\Rightarrow \quad x^{2}+1000 x-750 x-750000=0\\
    & \Rightarrow x(x+1000)-750(x+1000)=0\\
    &\Rightarrow (x+1000)(x-750) =0 \\
    &\Rightarrow x+1000=0 \text { or } x-750=0 \\
    &\Rightarrow x =-1000 \text { or } x=750\\
    &\Rightarrow x=750 & (\because \text{Speed cannot be negative})
    \end{aligned}
    \]

    Hence, the usual speed of plane is $750 \mathrm{km} / \mathrm{hr}$. In this question, pilot’s caring behaviour toward passengers is shown as well as other side pilot is cautious and alert for his duty to reach destination point at scheduled time.

    Question 4: A rectangular park is to be designed whose breadth is $3 \mathrm{m}$ less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude $12 \mathrm{m}$. Find the length and breadth of the rectangular park.

    Solution:


    Let length of rectangular park be $x$ m and breadth be $(x-3)$ m
    \[\begin{aligned}
    &\text { Base of isosceles } \Delta =(x-3) \mathrm{m} \\
    &\text { Altitude of } \Delta =12 \mathrm{m}
    \end{aligned}\]
    According to question,
    \[\begin{aligned}
    &\text{Area of Rectangular Park = Area of }\Delta +4\\
    &\Rightarrow x(x-3)=\frac{1}{2}(x-3) \times 12+4\\
    &\Rightarrow \quad x(x-3)=6(x-3)+4\\
    &\Rightarrow x^{2}-3 x=6 x-18+4\\
    &\Rightarrow x^{2}-9 x+14=0\\
    &\Rightarrow (x-7)(x-2)=0 \\
    &x=7 \text { or } x=2
    \end{aligned}
    \]
    But $x=2$ is rejected otherwise breadth will be -ve which is not possible.
    $\therefore$ Length of rectangular park is $7 \mathrm{m}$ and breadth is $4 \mathrm{m} .$

    Question 5: Solve the following quadratic equation for $x: 9 x^{2}-6 b^{2} x-\left(a^{4}-b^{4}\right)=0$

    Solution:
    The given quadratic equation can be written as $\left(9 x^{2}-6 b^{2} x+b^{4}\right)-a^{4}=0$
    $\Rightarrow\left(3 x-b^{2}\right)^{2}-\left(a^{2}\right)^{2}=0$
    $\Rightarrow\left(3 x-b^{2}+a^{2}\right)\left(3 x-b^{2}-a^{2}\right)=0 \left[\because x^{2}-y^{2}=(x+y)(x-y)\right]$
    $\Rightarrow 3 x-b^{2}+a^{2}=0$ or $3 x-b^{2}-a^{2}=0$
    $\Rightarrow 3 x=b^{2}-a^{2}$ or $3 x=b^{2}+a^{2}$
    \[\text{Hence, } x=\frac{b^{2}-a^{2}}{3}, \frac{b^{2}+a^{2}}{3}\]


    Solution of Quadratic equation by Completing the Square

    Question 1: Solve $(3 x+2)(2 x+3)=6$ by completing the square.
    Solution.
    \[\begin{aligned}
    &(3 x+2)(2 x+3)=6\\
    &\Rightarrow 6 x^{2}+13 x+6=6 \\
    &\Rightarrow 6 x^{2}+13 x=0 \\
    &\Rightarrow x^{2}+\frac{13}{6} x=0 \\
    &\Rightarrow x^{2}+\frac{13}{6} x+\left(\frac{13}{12}\right)^{2}-\left(\frac{13}{12}\right)^{2}=0 \\
    &\Rightarrow \quad\left(x+\frac{13}{12}\right)^{2}=\left(\frac{13}{12}\right)^{2} \\
    &\Rightarrow \quad x+\frac{13}{12}=\pm \frac{13}{12} \\
    &\Rightarrow x= -\frac{13}{12}+ \frac{13}{12},-\frac{13}{12}-\frac{13}{12}\\
    &\Rightarrow x= 0,-\frac{13}{6}
    \end{aligned}
    \]

    Question 2: Solved the equation $a\left(x^{2}+1\right)-x\left(a^{2}+1\right)=0$ by Quadratic formula.

    Solution.Given equation
    \[
    \begin{array}{r}
    a\left(x^{2}+1\right)-x\left(a^{2}+1\right)=0 \\
    \Rightarrow \quad a x^{2}-\left(a^{2}+1\right) x+a=0
    \end{array}
    \]
    Comparing with $A x^{2}+B x+C=0$
    \[
    A=a, B=-\left(a^{2}+1\right), C=a
    \]
    \[
    \begin{aligned}
    &\therefore x=\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A} \\
    &=\frac{-(-(a^{2}+1) ) \pm \sqrt{(-(a^{2}+1 ) )^{2}-4(a)(a)}}{2 a} \\
    &=\frac{a^{2}+1 \pm \sqrt{a^{4}+2 a^{2}+1-4 a^{2}}}{2 a} \\
    &=\frac{a^{2}+1 \pm \sqrt{a^{4}-2 a^{2}+1}}{2 a} \\
    &=\frac{a^{2}+1 \pm \sqrt{\left(a^{2}-1\right)^{2}}}{2 a} \\
    &=\frac{a^{2}+1 \pm\left(a^{2}-1\right)}{2 a} \\
    &=\frac{2 a^{2}}{2 a}, \frac{2}{2 a}=a, \frac{1}{a}
    \end{aligned}
    \]

    Question 3: The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal 2. 16/21 find the fraction .

    Solution:
    Let the fraction be $\frac{x}{2 x+1}$
    According to question,
    \[
    \frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21}
    \]
    Consider $\frac{x}{2 x+1}=y,$ then above equation becomes,
    \[
    \begin{aligned}
    & y+\frac{1}{y}=2 \frac{16}{21} \\
    & \Rightarrow \frac{y^{2}+1}{y}=\frac{58}{21} \\
    & \Rightarrow 21 y^{2}+21=58 y\\
    & \Rightarrow 21 y^{2}-58 y+21=0\\
    & \Rightarrow y=\frac{58 \pm \sqrt{58^{2}-4 \times 21 \times 21}}{2 \times 21}\\
    & \Rightarrow y=\frac{58 \pm \sqrt{3364-1764}}{42}\\
    & \Rightarrow y=\frac{58 \pm \sqrt{1600}}{42}\\
    & \Rightarrow y=\frac{58 \pm 40}{42}\\
    & \Rightarrow y=\frac{58+40}{42} \text{ or } y=\frac{58-40}{42}
    \end{aligned}
    \]


    \begin{aligned}
    &\Rightarrow \quad y=\frac{58+40}{42} \\
    &\Rightarrow \quad y=\frac{98}{42} \Rightarrow y=\frac{7}{3} \\
    &\Rightarrow \quad \frac{x}{2 x+1}=\frac{7}{3} \\
    &\Rightarrow \quad 3 x=14 x+7 \\
    &\Rightarrow \quad-11 x=7 \\
    &\Rightarrow \quad x=\frac{-7}{11} \text { (Rejected) }
    \end{aligned}

    \begin{aligned}
    &y= \frac{58-40}{42} \\
    &\Rightarrow y=\frac{18}{42} \Rightarrow y=\frac{3}{7} \\
    &\Rightarrow \frac{x}{2 x+1}=\frac{3}{7} \\
    &\Rightarrow 7 x=6 x+3 \\
    &\Rightarrow x=3 \\
    &\Rightarrow \frac{x}{2 x+1}=\frac{3}{2 \times 3+1}=\frac{3}{7}
    \end{aligned}

    Hence, the required fraction is 3/7.

    Question 4 :The sum of first $ n $ even natural numbers is given by the relation $ S = n (n + 1) $. Find the value of $ n $ if the sum is 420.

    Solution: According to Question,
    \[ \begin{aligned} n(n+1)=420 \\ \Rightarrow n^{2}+n=420 \\ \Rightarrow n^{2}+n-420 & =0 \end{aligned} \]
    Comparing with, $A n^{2}+B n+C=0$ \[ \begin{aligned} &A=1, B =1, C=-420 \\ &\therefore B^{2}-4 A C =(1)^{2}-4(1)(-420) \\ &=1+1680 \\ &=1681(>0) \end{aligned} \] $\therefore $The actual roots of the equation given using the quadratic formula are, \[ \begin{aligned} n =\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A} \ =\frac{-1 \pm \sqrt{(1)^{2}-4(1)(-420)}}{2(1)} \ =\frac{-1 \pm \sqrt{1681}}{2} \ =\frac{-1 \pm 41}{2} \ =\frac{-1+41}{2}, \frac{-1-41}{2} \ &=20,-21 \end{aligned} \] $n=-21$ is invalid because $n$ is a natural number \[ \therefore n=20 \] So,the value of $n$ is 20.

    Question 5: Solve by completing square method : $x^2−5x+5=0$

    Solution:
    $x^{2}-5 x+5=0$
    $x^{2}-5 x=-5$
    (1) Divide the equation by the coefficient of $\mathrm{x}^{2}$
    \[x^{2}-\frac{5}{1} x=-\frac{5}{1}\]
    (2) Add the square of half the coefficient of $x,$ both sides
    \[\begin{aligned}
    & x^{2}-5 x+\left(\frac{5}{2}\right)^{2}=-5+\left(\frac{5}{2}\right)^{2}\\
    & \left(x-\frac{5}{2}\right)^{2}=-5+\frac{25}{4}=\frac{5}{4}\\
    & \Rightarrow x-\frac{5}{2}=\pm \sqrt{\frac{5}{4}}=\pm \frac{\sqrt{5}}{2}\\
    & \mathrm{x}=\frac{5}{2} \pm \frac{\sqrt{5}}{2}
    \end{aligned} \]

    Nature of Roots

    Question 1 : If the roots of the quadratic equation $ x ^ {2} -2 k x-60 = 0 $ are equal, find the value of $ k $.

    Solution: Given quadratic equation,
    \[
    x^{2}-2 k x-6=0
    \]
    Comparing with $a x^{2}+b x+c=0$

    \[
    \begin{aligned}
    &a =1 \\
    &b =-2 k \\
    &c =-6 \\
    &\therefore\text { Discriminant } =b^{2}-4 a c\\
    &=(-2 k)^{2}-4(1)(-6) \\
    &=4 k^{2}+24
    \end{aligned}
    \]
    For equal roots,
    \[
    \begin{aligned}
    &\text{Discriminant }=0\\
    &\Rightarrow 4 k^{2}+24=0\\
    &\Rightarrow k^{2}+6=0 \quad [\text{dividing by 4}]\\
    &\Rightarrow k^{2}=-6
    \end{aligned}
    \]
    which is impossible $\because$ The square of any real number cannot be negative.

    Therefore no real value of $ k $ is possible.

    Question 2: Is the following situation possible? If so, determine their present ages.
    The sum of ages of two friends is 20 years. Four years ago, the product of their ages in years was $48 .$

    Solution:
    Let present age of one friend be $x$ years. So the sum of their ages being 20 years, the age of the second friend is $(20-x)$ years.
    Four years ago their respective ages were,
    for first friend: $(x-4)$ years, and,
    for second friend: $(20-x-4)=(16-x)$ years.
    The product of these two is,
    $(x-4)(16-x)=48,$ as per the problem
    Or, $-x^{2}+20 x-64=48$
    Or, $x^{2}-20 x+112=0$
    The coefficients are,
    $a=1, b=-20$ and $c=112$
    The value of the discriminant is,
    $b^{2}-4 a c=400-448=-48$
    The discriminant being negative, the given situation is not possible. The quadratic equation does not have any real roots.
    So the situation is not possible.

    Question 3:
    Is it possible to design a rectangular park of perimeter $80 \mathrm{m}$ and area $400 \mathrm{sq} \mathrm{m}$ ? If so, find its length and breadth.

    Solution :
    Let us assume the breadth of the park is $x$ m and length $y$ m. Its perimeter is then,
    $2(x+y)=80$
    $\Rightarrow x+y=40$
    $\Rightarrow y=40-x$
    Also the area is product of the length and breadth,
    $x(40-x)=400$
    $\Rightarrow -x^{2}+40 x=400$
    $\Rightarrow x^{2}-40 x+400=0$
    The coefficients are,
    $a=1, b=-40$ and $c=400$
    The discriminant is,
    $b^{2}-4 a c=1600-1600=0$
    Both the roots are then of equal real value.
    The root value is,
    Breadth $x=\frac{40}{2}=20 \mathrm{m},$ and
    Length $y=40-x=20 \mathrm{m}$
    Answer: Yes, it is possible. Both length and breadth are of equal value $20 \mathrm{m}$.

    Question 4. For which value of $k$ ,$(4-k) x^{2}-(2 k+4) x+(8 k+1)=0$ is a complete square.

    Solution.
    $a=4-k$
    \[
    \begin{array}{l}
    b=-(2 k+4) \\
    c=8 k+1
    \end{array}
    \]
    For complete square,
    \[
    b^{2}=4 a c
    \]
    \[
    \begin{array}{l}
    \Rightarrow \quad{-(2 k+4)}^{2}=4(4-k)(8 k+1) \\
    \Rightarrow \quad{-2(k+2)}^{2}=4(4-k)(8 k+1 \\
    \Rightarrow \quad 4(k+2)^{2}=4(4-k)(8 k+1) \\
    \Rightarrow \quad(k+2)^{2}=(4-k)(8 k+1) \\
    \Rightarrow \quad k^{2}+4 k+4=32 k+4-8 k^{2}-k

    \Rightarrow \quad 9 k^{2}-27 k=0\\
    \Rightarrow \quad k^{2}-3 k=0\\
    \Rightarrow \quad k(k-3)=0$\\

    \Rightarrow \quad k=0 \quad \text { or } \quad k-3=0\\

    \Rightarrow \quad k=0 \quad \text { or } \quad k=3\\
    \Rightarrow \quad k=0,3
    \end{array}
    \]

    Question 5: Discuss the nature of the roots of a quadratic equation $2 x^{2}-8 x+3=0$

    Solution: Here, the coefficients are all rational. The discriminant D for a given equation will be
    \[
    \begin{aligned}
    &D=b^{2}-4 a c=(-8)^{2}-4^{} 2^{} 3\\
    &=64-24\\
    &=40>0
    \end{aligned}
    \]
    We can see, the discriminant of the given quadratic equation is positive but not a perfect square. Hence, the roots of a quadratic equation are real, unequal and irrational.

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