10th Maths 4.4
NCERT Class 10th solution of Exercise 4.1
NCERT Class 10th solution of Exercise 4.2
NCERT Class 10th solution of Exercise 4.3
Exercise 4.4
Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them ;
i) `2x^2-3x+5=0`
ii) `3x^2-4sqrt3x+4=0`
iii) `2x^2-6x+3=0`
Sol. :
i) `2x^2-3x+5=0` compare with `ax^2+bx+c=0`
where `a=2, b=-3` and `c=5`
The Discriminant
`D =b^2-4ac`
`D =(-3)^2-4(2)(5)`
`D =9-40`
`D =-31<0`
no real roots.
Answer :
Real roots do not exist.
ii) `3x^2-4sqrt3x+4=0` compare with `ax^2+bx+c=0`
Where `a=3, b=4sqrt3` and `c=4`
The Discriminant
`D =b^2-4ac`
`D =(4sqrt3)^2-4(3)(4)`
`D =48-48`
`D =0`
two equal roots.
Now
`x=(-b+-sqrt(b^2-4ac))/2a`
`x=(-(4sqrt3)+-sqrt(b^2-4ac))/(2times3)`
`x=(4sqrt3+-0)/6`
`x=(4sqrt3)/6`
`x=(2sqrt3)/3`
`x=(2sqrt3)/(sqrt3timessqrt3)`
`x=2/sqrt3`
Answer :
Equal roots ; `2/sqrt3, 2/sqrt3`.
iii) `2x^2-6x+3=0` compare with `ax^2+bx+c=0`
Where `a=2, b=-6`, and `c=3`
The Discriminant D
`D =b^2-4ac`
`D =(-6)^2-4(2)(3)`
`D =36-24`
`D =12>0`
unequal real root.
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-(-6)+-sqrt(12))/(2(2))`
`x=(6+-sqrt(12))/4`
`x=(6+-2sqrt3)/4`
`x=(3+-sqrt3)/2`
Answer :
Unequal roots ; `(3+sqrt3)/2, (3-sqrt3)/2`.
Q2. Find the values of `k` for each of the following quadratic equations, so that they have two equal roots.
i) `2x^2+kx+3=0`
ii) `kx(x-2)+6=0`
Sol. :
i) `2x^2+kx+3=0` compare with `ax^2+bx+c=0`
Where `a=2,b=k`, and `c=3`
The roots are equal then
The Discriminant D `=0`
`D =b^2-4ac`
`0 =k^2-4(2)(3)`
`0 =k^2-24`
`k^2 =24`
`k =+-2sqrt6`
Answer :
`k=+-2sqrt6`.
ii) `kx(x-2)+6=0`
`kx^2-2kx+6=0` compare with `ax^2+bx+c=0`
Where `a=k, b=-2k` and `c=6`
The roots are equal then
The Discriminant D `=0`
`D =b^2-4ac`
`0=(-2k)^2-4(k)(6)`
`0=4k^2-24k`
`0=4k(k-6)`
`0=k-6`
`k=6`
Answer :
`k=6`.
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is `800m^2`? If so, find its length and breadth.
Sol. :
Let the breadth of the mango grove is `x` m,
and the length of the mango grove is `2x` m,
According to question
`x(2x)=800`
`2x^2=800`
`x^2=400`
`x=+-20`
negative breadth is not possible, so
breadth is `20` m.
length is `2(20)=40` m
Answer :
Yes, `40`m, `20` m.
Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is `20` years ago, the product of their ages in years was `48`.
Sol.
Let the present age of the first friend is `x` years
and the present age of the second friend is `(20-x)` years.
According to question
`(x-4)(20-x-4)=48`
`(x-4)(16-x)=48`
`x(16-x)-4(16-x)=48`
`16x-x^2-64+4x=48`
`-x^2+20x-64-48=0`
`x^2-20x+112=0` compare with `ax^2+bx+c=0`
Where `a=1, b=-20` and `c=112`
`D=b^2-4ac`
`D=(-20)^2-4(1)(112)`
`D=400-448`
`D=-48<0` so, given situation not possible.
Answer :
No, given situation not possible.
Q5. Is it possible to design a rectangular park of perimeter `80m` and area `400m^2`? If so, find its length and breadth.
Sol. :
Let the breadth of the park is `x` m.
and length of the park is `((perimeter)/2-breadth)`
`=80/2-x=(40-x)` m
According to question
`x(40-x)=400`
`40x-x^2=400`
`x^2-40x+400=0` compare with `ax^2+bx+c=0`
Where `a=1`, b=-40` and `c=400`
`D=b^2-4ac`
`D=(-40)^2-4(1)(400)`
`D=1600-1600`
`D=0` so, roots are equal.
`x=(-b+-sqrt(b^2-4ac))/2a`
`x=(-(-40)+-sqrt(0))/(2(1))`
`x=40/2=20` m
length `40-20=20` m
Answer :
Yes, length `20` m, breadth `20` m.
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