Quadratic Equations — Class 10 Maths Notes

Quadratic Equations · Class 10 Maths · 23 topics.

These notes are free to read without an account. Work through them in order, or use the chapter list to revise selectively before a test.

Topics covered in Quadratic Equations

  1. 1.Introduction of Quadratic Equations

    Quadratic equations are equations that look like 2++=0ax2+bx+c=0, where a, b, and c are numbers and x is the variable we're trying to find.

    1. Standard Form: 2++=0ax2+bx+c=0 is the standard form of a quadratic equation.
    2. Solving: To find the value of x, you can use either the Quadratic Formula or Factoring or Completing the Square methods.
    3. Quadratic Formula: =−±2−42x=2a−b±b2−4ac​​

    Real-Life Example

    Let's say you are trying to hit a target with a water balloon. The balloon's path is parabolic, resembling the graph of a quadratic equation. You want to find out at which points the balloon will hit the ground (these are the solutions x).

    Numerical Example

    For equation 2−3+2=0x2−3x+2=0 Using Quadratic Formula: =−(−3)±(−3)2−4×1×22×1x=2×1−(−3)±(−3)2−4×1×2​​ =3±9−82x=23±9−8​​ =3±12x=23±1​​ =3±12x=23±1​ =2,1x=2,1

  2. 2.Sam and Rahul together have 90 marbles. Both of them lost 10 marbles each, and the product of the number of marbles they now have is 248. We would like to find out how many marbles they had to start with.

    1. According to the first condition, Sam and Rahul together have 90 marbles. So, +=90S+R=90.

    2. Both lost 10 marbles each, so now Sam has −10S−10 marbles and Rahul has −10R−10 marbles.

    3. The product of the number of marbles they now have is 248, which gives us (−10)×(−10)=248(S−10)×(R−10)=248.

    We need to find S and R.

    Step 1: Solve the First Equation

    We already have +=90S+R=90.

    Step 2: Solve the Second Equation

    (−10)×(−10)=248(S−10)×(R−10)=248
    So, −10−10+100=248SR−10S−10R+100=248
    Or, −10−10=148SR−10S−10R=148

    Step 3: Use Both Equations to Solve for S and R

    We know +=90S+R=90, so =90−R=90−S
    Plug this into −10−10=148SR−10S−10R=148
    So, (90−)−10−10(90−)=148S(90−S)−10S−10(90−S)=148
    Solve this, you get 2−80+742=0S2−80S+742=0

    Now use the Quadratic Formula to solve for S: =80±802−4×1×7422S=280±802−4×1×742​​
    =80±2562S=280±256​​
    =80±162S=280±16​
    So, =32S=32 or 4848

    If =32S=32, then =90−32=58R=90−32=58
    If =48S=48, then =90−48=42R=90−48=42

  3. 3.The area of a rectangular plot is 528 m² , The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

    Given Situation

    • The area of the plot is 528 2528m2.
    • The length is one more than twice its breadth.

    Step 1: Define Variables

    • Let b be the breadth in meters.
    • Then, =2+1l=2b+1 meters will be the length.

    Step 2: Form Equation

    The formula for area is =×A=l×b.
    So, we get 528=(2+1)×528=(2b+1)×b.
    Expanding, we get 22+−528=02b2+b−528=0.

    Step 3: Solve Equation

    Factor the equation 22+−528=02b2+b−528=0 to (2−33)(+16)=0(2b−33)(b+16)=0.
    We have two possible values for b:
    =332b=233​ or =−16b=−16.
    Since breadth can't be negative, =332b=233​ or =16.5 b=16.5m.

    Step 4: Find Length

    =2+1l=2b+1
    Substitute =16.5b=16.5 into the equation:
    =2×16.5+1l=2×16.5+1
    =33+1l=33+1
    =34 l=34m

    Conclusion

    The breadth is 16.5 16.5m and the length is 34 34m.

  4. 4.The product of two consecutive positive integers is 306. We need to find the integers.

    1. Step 1: Understand the Problem

      • The product of two consecutive integers is 306.

    2. Step 2: Form the Equation

      • Let the first integer be x.
      • The next consecutive integer will be +1x+1.
      • Product = ×(+1)x×(x+1)
      • 306=×(+1)306=x×(x+1)

    3. Step 3: Write the Quadratic Equation

      • 306=2+306=x2+x
      • 2+−306=0x2+x−306=0

    4. Step 4: Solve for x

      • Factor the equation: (−17)(+18)=0(x−17)(x+18)=0
      • =17x=17 or =−18x=−18
      • Since we're looking for positive integers, =17x=17.
      • The next integer is +1=18x+1=18.

  5. 5.Rahul’s mother is 36 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rahul’s present age.

    1. Step 1: Understand the Problem

      • Rahul's mother is 36 years older than him.
      • The product of their ages 3 years from now will be 360.

    2. Step 2: Form the Equation

      • Let Rahul's current age be x years.
      • His mother's current age will be +36x+36 years.
      • Their ages 3 years from now will be +3x+3 and +36+3x+36+3.
      • Product of their future ages = (+3)(+39)=360(x+3)(x+39)=360

    3. Step 3: Write the Quadratic Equation

      • 2+42+117=360x2+42x+117=360
      • 2+42−243=0x2+42x−243=0

    4. Step 4: Solve for x

      • Factor the equation: (−9)(+27)=0(x−9)(x+27)=0
      • =9x=9 or =−27x=−27
      • Since age can't be negative, Rahul's current age is 9 years.

  6. 6.A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

    1. Step 1: Understand the Problem

      • The train travels a distance of 480 km.
      • If the speed were 8 km/h less, it would take 3 hours more.

    2. Step 2: Form the Equation

      • Let the original speed of the train be x km/h.
      • Original time taken = 480x480​ hours.
      • New speed = −8x−8 km/h.
      • New time taken = 480−8x−8480​ hours.
      • 480−8=480+3x−8480​=x480​+3

    3. Step 3: Write the Quadratic Equation

      • 480=480(−8)+3(−8)480x=480(x−8)+3x(x−8)
      • 480=480−3840+32−24480x=480x−3840+3x2−24x
      • 32−24−3840=03x2−24x−3840=0

    4. Step 4: Solve for x

      • Factor the equation: (3+60)(−64)=0(3x+60)(x−64)=0
      • =−60x=−60 or =64x=64
      • Since speed can't be negative, the original speed of the train is 6464 km/h.

  7. 7.Check whether the following are quadratic equations or not:

    1. 2+3+2=01. x2+3x+2=0

      • Is it a Quadratic Equation? Yes.
      • Why? Because it's in the form 2++=0ax2+bx+c=0.

    2. 2−3=7
      2. 2x−3=7

      • Is it a Quadratic Equation? No.
      • Why? Because the highest power of x is 1.

    3. 42=16
      3. 4x2=16

      • Is it a Quadratic Equation? Yes.
      • Why? Because it can be rewritten as 42−16=04x2−16=0.

    4. 52−33=0
      4. 5x2−3x3=0

      • Is it a Quadratic Equation? No.
      • Why? Because the highest power of x is 3.

  8. 8.Check whether the following are quadratic equations :-

    i.) x3−4x2−x+1=(x−2)3

    Solve: Step 1: Expand (−2)3(x−2)3

    (−2)3=3−62+12−8(x−2)3=x3−6x2+12x−8

    Step 2: Substitute into the Equation

    Substitute (−2)3(x−2)3 into 3−42−+1=(−2)3x3−4x2−x+1=(x−2)3: 3−42−+1=3−62+12−8x3−4x2−x+1=x3−6x2+12x−8

    Step 3: Simplify

    3−42−+1−3+62−12+8=0x3−4x2−x+1−x3+6x2−12x+8=0 22+11−7=02x2+11x−7=0

    Notice that after simplifying, we get a quadratic equation 22+11−7=02x2+11x−7=0.

    Step 4: Solve the Quadratic Equation

    To solve 22+11−7=02x2+11x−7=0, you can use factoring, completing the square, or the quadratic formula. Here, factoring won't work easily, so let's use the quadratic formula.

    =−±2−42x=2a−b±b2−4ac​​ Where =2,=11,=−7a=2,b=11,c=−7 =−11±121+564x=4−11±121+56​​ =−11±1774x=4−11±177​​

    So, the roots are: =−11+1774x=4−11+177​​ =−11−1774x=4−11−177​​

    Conclusion

    The roots of the equation 3−42−+1=(−2)3x3−4x2−x+1=(x−2)3 are −11+17744−11+177​​ and −11−17744−11−177​​.

  9. 9.Factorization Method to Solve Quadratic Equation :

      A quadratic equation is an equation of the form ax2+bx+c=0 a, b & c are constants.
    1. Factorization Step 1: Write down the equation 2++=0ax2+bx+c=0.
    2. Step 2: Make sure the equation is in standard form.
    3. Step 3: Factorize 2++ax2+bx+c into (+)(+)(px+q) (rx+s).
    4. Step 4: Set each factor equal to zero: +=0px+q=0 and +=0rx+s=0.
    5. Step 5: Solve for x in each equation.

    Example:

    Let's say we have 2−5+6=0x2−5x+6=0.

    1. Step 1: The equation is already in standard form.
    2. Step 2: Factorize 2−5+6x2−5x+6 into (−2)(−3)(x−2)(x−3).
    3. Step 3: Set −2=0x−2=0 and −3=0x−3=0.
    4. Step 4: Solve for x, =2x=2 or =3x=3. Real Life Example: Imagine you are trying to find out how many students are in two different classes. You know that the total number of students in both classes is 5 and the product of the number of students in each class is 6. This can be represented as 2−5+6=0x2−5x+6=0. By solving this equation, you find out that there are 2 students in one class and 3 in the other

  10. 10.Find the roots of the equation by Factorization method.

    1. 2x2−5x+3=0


    Solve:

    1. Step 1: Write down the equation 22−5+3=02x2−5x+3=0.
    2. Step 2: Factorize the equation. To do this, look for two numbers that add up to −5−5 and multiply to 2×3=62×3=6. The numbers are −3−3 and −2−2.
    3. Step 3: Rewrite the equation as 22−3−2+3=02x2−3x−2x+3=0.
    4. Step 4: Group the terms and factorize: (22−3)+(−2+3)=0(2x2−3x)+(−2x+3)=0.
    5. Step 5: Factor out the common terms: (2−3)−1(2−3)=0x(2x−3)−1(2x−3)=0.
    6. Step 6: Factorize further: (−1)(2−3)=0(x−1)(2x−3)=0.
    7. Step 7: Solve for x, =1x=1 or =32x=23​.
      8. 32−26+2=0
      2. 3x2−26​x+2=0
      1. Solve: Step 1: Write down the equation 32−26+2=03x2−26​x+2=0.
      2. Step 2: Factorize the equation. Look for two numbers that add up to −26−26​ and multiply to 3×2=63×2=6. The numbers are −6−6​ and −6−6​.
      3. Step 3: Rewrite the equation as 32−6−6+2=03x2−6​x−6​x+2=0.
      4. Step 4: Group the terms and factorize: (32−6)+(−6+2)=0(3x2−6​x)+(−6​x+2)=0.
      5. Step 5: Factor out the common terms: (3−6)−1(3−6)=0x(3x−6​)−1(3x−6​)=0.
      6. Step 6: Factorize further: (−1)(3−6)=0(x−1)(3x−6​)=0.
      7. Step 7: Solve for x, =1x=1 or =63x=36​​.

  11. 11.Find two numbers whose sum is 27 and product is 182.

    To find two numbers whose sum is 27 and product is 182, you can set up a quadratic equation and solve it using the factorization method.

    1. Step 1: Let the two numbers be x and 27−27−x.
    2. Step 2: Their product is 182, so ×(27−)=182x×(27−x)=182.
    3. Step 3: Expand and simplify to get 2−27+182=0x2−27x+182=0.
    4. Step 4: Factorize the equation into (−14)(−13)=0(x−14)(x−13)=0.
    5. Step 5: Solve for x, =14x=14 or =13x=13.

    So the two numbers are 14 and 13.

  12. 12.Find two consecutive positive integers, sum of whose squares is 365.

    To find two consecutive positive integers whose squares sum up to 365, you can set up a quadratic equation and solve it using the factorization method.

    1. Step 1: Let the two consecutive integers be x and +1x+1.
    2. Step 2: The sum of their squares is 365, so 2+(+1)2=365x2+(x+1)2=365.
    3. Step 3: Expand and simplify to get 2+2+2+1=365x2+x2+2x+1=365.
    4. Step 4: Combine like terms to get 22+2−364=02x2+2x−364=0.
    5. Step 5: Divide by 2 to simplify: 2+−182=0x2+x−182=0.
    6. Step 6: Factorize the equation into (−13)(+14)=0(x−13)(x+14)=0.
    7. Step 7: Solve for x, =13x=13 or =−14x=−14.

    Since we are looking for positive integers, =13x=13 and the next integer is +1=14x+1=14.

  13. 13.The altitude of a right triangle is 17 cm less than its base. If the hypotenuse is 26 cm, find the other two sides.

    Problem Statement

    We have a right triangle with:

    • Hypotenuse = 26 cm
    • Altitude = Base - 17 cm

    We need to find the lengths of the base and the altitude.

    Solve

    Step 1: Define the Variables

    Let the base of the triangle be x cm. Then, the altitude will be −17x−17 cm.

    Step 2: Use the Pythagorean Theorem

    In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, we have:

    262=2+(−17)2262=x2+(x−17)2

    Step 3: Expand and Simplify

    First, square 26 to get 676. Then expand (−17)2(x−17)2 to 2−34+289x2−34x+289.

    So, the equation becomes: 676=2+2−34+289676=x2+x2−34x+289

    Combine like terms: 676=22−34+289676=2x2−34x+289

    Step 4: Move All Terms to One Side

    Move all terms to one side to set the equation to zero: 0=22−34−3870=2x2−34x−387

    Step 5: Factorize the Equation

    Divide every term by 2: 0=2−17−193.50=x2−17x−193.5

    Factorize this equation into: 0=(−22.5)(+8.5)0=(x−22.5)(x+8.5)

    Step 6: Solve for x

    Now, we have two possible solutions for x:

    1. −22.5=0i) x−22.5=0 => =22.5x=22.5 cm
    2. +8.5=0ii) x+8.5=0 => =−8.5x=−8.5 cm

    Since length cannot be negative, we take =22.5x=22.5 cm as the base.

    Step 7: Find the Altitude

    The altitude will be −17=22.5−17=5.5x−17=22.5−17=5.5 cm.

    So, the base is 22.5 cm and the altitude is 5.5 cm.

  14. 14.Nature of Roots

    When you have a quadratic equation, which looks like 2++=0ax2+bx+c=0, you can find its roots. Roots are the values of x that make the equation true. The nature of roots tells us what kind of roots we will get. Will they be real numbers? Will they be imaginary? Will they be the same or different?

    You can find out the nature of roots by using the discriminant, which is 2−4b2−4ac.

    1. Real and Different Roots: If 2−4>0b2−4ac>0, then you get two different real roots.
    2. Real and Same Roots: If 2−4=0b2−4ac=0, then you get one real root (or two same real roots).
    3. Imaginary Roots: If 2−4<0b2−4ac<0, then you get two imaginary roots.

    Real-life Example:

    Imagine you are throwing a ball up in the air. The height of the ball at different times can be represented by a quadratic equation. The roots of this equation tell you when the ball will hit the ground. If you get real roots, it means the ball will hit the ground at those times. If you get imaginary roots, it's like saying the ball will never hit the ground, which is not possible in real life for a thrown ball.

  15. 15.Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots

    Certainly! Let's find the discriminant and the nature of the roots for the quadratic equation 22−4+3=02x2−4x+3=0.

    The formula for the discriminant is

    2−4b2−4ac.

    Here, =2a=2, =−4b=−4, and =3c=3.

    So, the discriminant will be (−4)2−4×2×3(−4)2−4×2×3.

    That is 16−2416−24.

    So, the discriminant is −8−8.

    Because the discriminant is negative (−8<0−8<0), the roots of the equation will be imaginary.

    Real-life Example:

    Imagine you're trying to solve a puzzle, and the quadratic equation represents the possible solutions. A negative discriminant is like saying there's no "real-world" solution to the puzzle, only imaginary or theoretical ones.

  16. 16.A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? I

    Let's assume the distance from gate A to the pole is x meters. Then, the distance from gate B to the pole would be (13−)(13−x) meters, because the diameter of the park is 13 meters.

    The difference in distances is given as 7 meters. So, we can write the equation as:

    ∣−(13−)∣=7∣x−(13−x)∣=7 ∣2−13∣=7∣2x−13∣=7 (2−13)2=49(2x−13)2=49 42−52+169=494x2−52x+169=49 42−52+120=04x2−52x+120=0

    Now, let's find the discriminant to check the nature of the roots:

    2−4=(−52)2−4×4×120b2−4ac=(−52)2−4×4×120 2704−19202704−1920 784784

    The discriminant is 784, which is greater than 0. This means there are two real and distinct roots for x, so yes, it is possible to erect the pole at two different points.

    Real-life Example:

    Think of this like setting up a tent pole in a circular park. You have two gates, and you want to place the pole so that it's closer to one gate than the other by 7 meters. The math shows you can do it in two different spots!

  17. 17.Find the discriminant of the equation 3x²−2x+ 1/3=0. and hence find the nature of its roots. Find them, if they are real.

    Let's find the discriminant and the nature of the roots for the quadratic equation

    3x2−2x+31​=0.

    The formula for the discriminant is

    2−4b2−4ac.

    Here, =3a=3, =−2b=−2, and =13c=31​.

    So, the discriminant will be (−2)2−4×3×13(−2)2−4×3×31​.

    That is 4−44−4.

    So, the discriminant is 00.

    Because the discriminant is zero (00), the roots of the equation will be real and the same.

    To find the roots, you can use the quadratic formula =−±2−42x=2a−b±b2−4ac​​.

    Here, =−(−2)±02×3x=2×3−(−2)±0​​, =2±06x=62±0​, =26x=62​, =13x=31​.

    So, the root is =13x=31​, and it's real and the same.

    Real-life Example:

    Imagine you're trying to find the best price to sell lemonade. If the discriminant was zero, it would mean there's only one best price to maximize your profit.

  18. 18.Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

    22−3+5=01. 2x2−3x+5=0
    The formula for the discriminant is

    2−4b2−4ac.

    Here, =2a=2, =−3b=−3, and =5c=5.

    So, the discriminant will be (−3)2−4×2×5(−3)2−4×2×5.

    That is 9−409−40.

    So, the discriminant is −31−31.

    Because the discriminant is negative (−31<0−31<0), the roots of the equation will be imaginary.

    2. 3x2−43​x+4=0

    Here, =3a=3, =−43b=−43​, and =4c=4.

    The discriminant will be (−43)2−4×3×4(−43​)2−4×3×4.

    That is 48−4848−48.

    So, the discriminant is 00.

    Because the discriminant is zero (00), the roots of the equation will be real and the same.

    To find the roots, you can use the quadratic formula =−±2−42x=2a−b±b2−4ac​​.

    Here, =−(−43)±02×3x=2×3−(−43​)±0​​, =436x=643​​, =233x=323​​.

    So, the root is =233x=323​​, and it's real and the same.

  19. 19.Find the values of k for each of the following quadratic equations, so that they have two equal roots.

    First Equation: 22++3=02x2+kx+3=0

    Here,

    =2a=2, =b=k, and =3c=3.

    The discriminant will be 2−4×2×3k2−4×2×3.

    Setting it to zero, we get: 2−24=0k2−24=0 2=24k2=24 =±24k=±24​ =±26k=±26​

    So, the value of k can be 2626​ or −26−26​ for the equation to have two equal roots.

    Second Equation: (−2)+6=0kx(x−2)+6=0

    Solve:

    Expanding the equation, we get 2−2+6=0kx2−2kx+6=0.

    Here, =a=k, =−2b=−2k, and =6c=6.

    The discriminant will be (−2)2−4××6(−2k)2−4×k×6.

    Setting it to zero, we get: 42−24=04k2−24k=0 4(−6)=04k(k−6)=0 k = 0 \) or \( k = 6

    So, the value of k can be 00 or 66 for the equation to have two equal roots.

  20. 20.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.

    Let the breadth of the mango grove be x meters. Then, the length will be 22x meters.

    Area = ×2=800x×2x=800 22=8002x2=800 2=400x2=400 2−400=0x2−400=0

    Here, =1a=1, =0b=0, =−400c=−400.

    The discriminant 2−4=02−4×1×(−400)=1600b2−4ac=02−4×1×(−400)=1600.

    Since the discriminant is positive, it is possible to design such a rectangular mango grove.

    To find x, =400x=400​ =20x=20

    So, the breadth is 2020 m and the length is 4040 m.

  21. 21.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.

    Let the age of one friend be x years. Then, the age of the other friend will be 20−20−x years.

    Four years ago, their ages were −4x−4 and 20−−420−x−4.

    The product of their ages four years ago was 48: (−4)(20−−4)=48(x−4)(20−x−4)=48 2−16+48=0x2−16x+48=0

    Here, =1a=1, =−16b=−16, =48c=48.

    The discriminant 2−4=(−16)2−4×1×48=256−192=64b2−4ac=(−16)2−4×1×48=256−192=64.

    Since the discriminant is positive, it is possible for such ages to exist.

    To find x, we can use the quadratic formula or factorization. Here, =8x=8 or =12x=12.

    So, the ages could be 88 and 1212 years or vice versa.

  22. 22.Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

    Let the breadth be x meters. Then, the length will be 40−40−x meters (because the perimeter is 80, and 2+2(40−)=802x+2(40−x)=80).

    Area = ×(40−)=400x×(40−x)=400 2−40+400=0x2−40x+400=0

    Here, =1a=1, =−40b=−40, =400c=400.

    The discriminant 2−4=(−40)2−4×1×400=1600−1600=0b2−4ac=(−40)2−4×1×400=1600−1600=0.

    Since the discriminant is zero, it is possible to design such a rectangular park.

    To find x, =20x=20.

    So, the breadth is 2020 m and the length is 2020 m.

  23. 23.Quick Revision

    1. Introduction of Quadratic Equations: A quadratic equation is a type of polynomial that always has the highest power of its variable as 2. It looks like 2++=0ax2+bx+c=0, where a, b, and c are numbers and a is not zero.

    2. Factorization Method to Solve Quadratic Equation: The factorization method involves writing the quadratic as the product of two brackets. For example, 2+5+6=0x2+5x+6=0 can be factored into (+2)(+3)=0(x+2)(x+3)=0. You then find the values of x that make each bracket zero.

    3. Find the Roots of the Equation by Factorization Method: To find the roots (solutions) of the quadratic equation by factorization, follow these steps:

    • Express the equation in the form of 2++=0ax2+bx+c=0.
    • Factorize it into two brackets.
    • Set each bracket equal to zero and solve for x.

    4. Nature of Roots: The nature of the roots of a quadratic equation depends on the discriminant (2−4b2−4ac):

    • If the discriminant is positive, you get two real and distinct roots.
    • If it's zero, you get exactly one real root (repeated).
    • If it's negative, the roots are complex and not real.

    Important Formulas and Implementation

    • Quadratic Formula: =−±2−42x=2a−b±b2−4ac​​. Use this when factorization is difficult.
    • Discriminant: =2−4D=b2−4ac. This tells you the nature of the roots.

More Class 10 Maths chapters