calendarBack
Quant

/

Algebra

/

Quadratic Equations
ALL MODULES

CAT 2025 Lesson : Quadratic Equations - Other Types

bookmarked

4.6 Other Types

Example 20

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

Solution

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

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

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

If a=1a = 1 and d=10d = 10, then the other 22 terms in AP are b=4b = 4 and c=7c = 7. Here, bc=28bc = 28 (Accepted)

If a=2a = 2 and d=5d = 5, then the other 22 terms in AP are b=3b = 3 and c=4c = 4. Here, bc=12bc = 12 (Rejected)

∴ a=1,b=4,c=7a = 1, b = 4, c = 7 and d=10d = 10 are the 44 roots.

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

Answer: −22-22


Example 21

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

Solution

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

⇒ x=y2=2z x = \dfrac{y}{2} = 2z

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

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

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

Answer: 2727


Example 22

Some kids were playing with marbles, and they realised that the average number of marbles with each kid was 44 more than the number of kids. If the total marbles was 320320, then what is the number of kids?

Solution

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

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

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

Alternatively

As the average and number of kids are integers, we need to express 320320 as a product of 22 numbers, such that their difference is 44.

320=20×16320 = 20 \times 16

As the average is 44 more than the number of students, 2020 is the average and 16 is the number of students.

Answer: 1616


Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock