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Quadratic Equations

Quadratic Equations

MODULES

Basics of Polynomial & Quadratic Equations
Discriminant & Graphical Representation
Sum & Product of Roots
Factorisation Method
Formulation & Completion of Squares
Changes to Roots
Mistakes in Roots, Common Roots & Squaring
Infinite Series & Transposed
Other Types
Higher Order Equations
Synthetic Division & Remainder Theorem
Maxima, Minima & Descrates Rule
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Quadratic Equations 1
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Quadratic Equations 2
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Quadratic Equations 3
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PRACTICE

Quadratic Equations : Level 1
Quadratic Equations : level 2
Quadratic Equations : level 3
ALL MODULES

CAT 2025 Lesson : Quadratic Equations - Other Types

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4.6 Other Types

Example 20

a,b,ca, b, ca,b,c and ddd are positive integers that are in increasing Arithmetic Progression. If aaa and ddd are the roots of x2+mx+10=0x^{2} + mx + 10 = 0x2+mx+10=0, while bbb and ccc are the roots of x2+nx+28=0x^{2} + nx + 28 = 0x2+nx+28=0, then m+n=m + n =m+n= ?

Solution

Product of roots are known in both the equations.
⇒ ad=10 ad = \bm{10}ad=10 and bc=28bc = \bm{28}bc=28

As a,b,ca, b, ca,b,c and ddd are positive integers in AP, we start by writing adadad as product of integers.

ad=1×10=2×5ad = 1 \times 10 = 2 \times 5ad=1×10=2×5

If a=1a = 1a=1 and d=10d = 10d=10, then the other 222 terms in AP are b=4b = 4b=4 and c=7c = 7c=7. Here, bc=28bc = 28bc=28 (Accepted)

If a=2a = 2a=2 and d=5d = 5d=5, then the other 222 terms in AP are b=3b = 3b=3 and c=4c = 4c=4. Here, bc=12bc = 12bc=12 (Rejected)

∴ a=1,b=4,c=7a = 1, b = 4, c = 7a=1,b=4,c=7 and d=10d = 10d=10 are the 444 roots.

m=−(a+d)=−11m = -(a + d) = -11m=−(a+d)=−11 and n=−(b+c)=−11n = -(b + c) = -11n=−(b+c)=−11
∴ m+n=−22m + n = -22m+n=−22

Answer: −22-22−22


Example 21

If 3x=9(y4)=27(2z3)3^{x} = 9^{(\frac{y}{4})} = 27^{(\frac{2z}{3})}3x=9(4y​)=27(32z​) and xyz=64xyz = 64xyz=64, then x+2y+7=x + 2y + 7 =x+2y+7= ?

Solution

3x3^{x}3x =9(y4)= 9^{(\frac{y}{4})}=9(4y​) =27(2z3)= 27^{(\frac{2z}{3})}=27(32z​)    ⇒    3x=3(2×y4)3^{x} = 3^{(2 \times \frac{y}{4})}3x=3(2×4y​)    ⇒    3(3×2z3)3^{(3 \times \frac{2z}{3})}3(3×32z​)    ⇒    3x=3y2=32z 3^{x} = 3^{\frac{y}{2}} = 3^{2z}3x=32y​=32z

⇒ x=y2=2z x = \dfrac{y}{2} = 2zx=2y​=2z

⇒ y=2x y = 2xy=2x and z=x2z = \dfrac{x}{2}z=2x​

As xyz=64xyz = 64xyz=64    ⇒    x×2x×x2=64 x \times 2 x \times \dfrac{x}{2} = 64x×2x×2x​=64    ⇒    x=4 x = 4x=4 and y=8y = 8y=8

x+2y+7=4+2×8+7=27x + 2y + 7 = 4 + 2 \times 8 + 7 = 27x+2y+7=4+2×8+7=27

Answer: 272727


Example 22

Some kids were playing with marbles, and they realised that the average number of marbles with each kid was 444 more than the number of kids. If the total marbles was 320320320, then what is the number of kids?

Solution

Let the number of kids be xxx.
∴ Average number of marbles =x+4= x + 4=x+4

x(x+4)=320x(x + 4) = 320x(x+4)=320
⇒ x2+4x−320=0 x^{2} + 4x - 320 = 0x2+4x−320=0
⇒ (x+20)(x−16)=0 (x + 20) (x - 16) = 0(x+20)(x−16)=0
⇒ x=16,−20 x = 16, -20x=16,−20

As xxx cannot be negative, number of kids =16= \bm{16}=16

Alternatively

As the average and number of kids are integers, we need to express 320320320 as a product of 222 numbers, such that their difference is 444.

320=20×16320 = 20 \times 16320=20×16

As the average is 444 more than the number of students, 202020 is the average and 16 is the number of students.

Answer: 161616


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