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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 - Mistakes in Roots, Common Roots & Squaring

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4. Common Types

4.1 Mistake in Roots

Example 12

Simha formed a quadratic equation x2+ax+b=0x^{2} + ax + b = 0x2+ax+b=0. He noted when he changed one of the roots from −7-7−7 to +3+3+3, he got the equation x2−ax+c=0x^{2} - ax + c = 0x2−ax+c=0. Then, b=b =b= ?

(1) −10-10−10            (2) −14-14−14            (3) 101010            (4) 141414           

Solution

Let the initial roots of the equation be −7-7−7 and kkk.
From the equation, we get sum of roots =−7+k=−a⟶(1)= -7 + k = -a \longrightarrow (1)=−7+k=−a⟶(1)

After change, the roots of the equation are +3+3+3 and kkk.
From the changed equation, sum of roots =3+k=a⟶(2)= 3 + k = a \longrightarrow (2)=3+k=a⟶(2)

(1)+(2)(1) + (2) (1)+(2) ⇒ −4+2k=0 -4 + 2k = 0−4+2k=0
⇒ k=2 \bm{k = 2}k=2

bbb is the product of roots in the initial equation, where roots are −7-7−7 and kkk.
∴ b=−7k=−7×2=−14b = -7k = -7 \times 2 = -14b=−7k=−7×2=−14

Answer: (2) −14(2) \space -14(2) −14


Example 13

A student was asked to form a quadratic equation with roots aaa and bbb. Instead, she reversed the signs of the roots (positive to negative and vice-versa) and obtained the equation as 8x2−2x−1=08x^{2} - 2x - 1 = 08x2−2x−1=0. Find the actual quadratic equation with roots aaa and bbb?

Solution

8x2−2x−1=08x^{2} - 2x - 1 = 08x2−2x−1=0
⇒ 8x2−4x+2x−1=0 8x^{2} - 4x + 2x - 1 = 08x2−4x+2x−1=0
⇒ 4x(2x−1)+1(2x−1)=04x (2x - 1) + 1(2x - 1) = 04x(2x−1)+1(2x−1)=0
⇒ (4x+1)(2x−1)=0(4x + 1) (2x - 1) = 0(4x+1)(2x−1)=0

x=−14,12x = \dfrac{-1}{4}, \dfrac{1}{2}x=4−1​,21​

As the signs of the roots were reversed, roots of the actual quadratic equation should be x=14,−12x = \dfrac{1}{4}, \dfrac{-1}{2}x=41​,2−1​

Quadratic Equation =(x−14)(x+12)=0= \left(x - \dfrac{1}{4} \right) \left(x + \dfrac{1}{2} \right) = 0=(x−41​)(x+21​)=0

=(x−14)(x+12)=0= \left(x - \dfrac{1}{4} \right) \left(x + \dfrac{1}{2} \right) = 0=(x−41​)(x+21​)=0

⇒ (4x−14)(2x+12)=0 \left( \dfrac{4x - 1}{4} \right) \left(\dfrac{2x + 1}{2} \right) = 0(44x−1​)(22x+1​)=0 ⇒ (4x−1)(2x+1)=0 (4x - 1)(2x + 1) = 0(4x−1)(2x+1)=0

⇒ 8x2+2x−1=0 8x^{2} + 2x - 1 = 08x2+2x−1=0

Answer: 8x2+2x−1=08x^{2} + 2x - 1 = 08x2+2x−1=0


4.2 Common Roots

In case two equations are said to have common roots, then equate the two equations and solve them. Substitute the solutions in the two initial equations to verify if they are roots.

Example 14

What are the common roots for the equations
(I) x2−3x+2=0x^{2} - 3x + 2 = 0x2−3x+2=0 and 3x2−5x−2=03x^{2} - 5x - 2 = 03x2−5x−2=0 ?
(II) x2−3x+2=0x^{2} - 3x + 2 = 0x2−3x+2=0 and x2−x−4=0x^{2} - x - 4 = 0x2−x−4=0 ?

Solution

When 222 equation are said to have common roots, we can equate and solve them.

Case I: x2−3x+2=3x2−5x−2x^{2} - 3x + 2 = 3x^{2} - 5x - 2x2−3x+2=3x2−5x−2
⇒ 2x2−2x−4=0 2x^{2} - 2x - 4 = 02x2−2x−4=0 (−2-2−2 can be split as −4-4−4 and 222)
x=42,−22=2,−1x = \dfrac{4}{2}, \dfrac{-2}{2} = 2, -1x=24​,2−2​=2,−1

At x=2,x2−3x+2=0\bm{x = 2}, x^{2} - 3x +2 = 0x=2,x2−3x+2=0, which is, therefore, a common root.
At x=−1x = -1x=−1, x2−3x+2=6x^{2} - 3x + 2 = 6x2−3x+2=6, which is not a common root.

Case II: x2−3x+2=x2−x−4x^{2} - 3x + 2 = x^{2} - x - 4x2−3x+2=x2−x−4
⇒ 2x=6 2x = 62x=6
⇒ x=3x = 3x=3

At x=3,x2−3x+2=2x = 3, x^{2} - 3x + 2 = 2x=3,x2−3x+2=2. Therefore, this is not a common root.

Answer: (I) 222; (II) No common root


4.3 Squaring Equations

These questions will have square roots of linear or quadratic expressions. We need to repeatedly square these equations to remove the square roots before proceeding solve them.

Example 15

Solve for xxx when (x+1)+(x−2)=(5−x)\sqrt{(x + 1)} + \sqrt{(x - 2)} = \sqrt{(5 - x)}(x+1)​+(x−2)​=(5−x)​ and all three square roots result in non-negative real numbers.

Solution

Squaring both sides,
(x+1)+(x−2)2=5−x\sqrt{(x + 1)} + \sqrt{(x - 2)}^{2} = 5 - x(x+1)​+(x−2)​2=5−x
⇒ x+1+x−2+2(x+1)(x−2)=5−x x + 1 + x - 2 + 2 \sqrt{(x + 1) (x - 2)} = 5 - xx+1+x−2+2(x+1)(x−2)​=5−x
⇒ 2x2−x−2=6−3x 2 \sqrt{x^{2} - x - 2} = 6 - 3x2x2−x−2​=6−3x

Squaring both sides,
⇒ 4(x2−x−2)=36+9x2−36x 4 (x^{2} - x - 2) = 36 + 9x^{2} - 36x4(x2−x−2)=36+9x2−36x
⇒ 5x2−32x+44=0 5x^{2} - 32x + 44 = 05x2−32x+44=0
⇒ 5x2−10x−22x+44=05x^{2} - 10x - 22x + 44 = 05x2−10x−22x+44=0
⇒ 5x(x−2)−22(x−2)=0 5x(x - 2) - 22(x - 2) = 05x(x−2)−22(x−2)=0
⇒ (5x−22)(x−2)=0(5x - 22)(x - 2) = 0(5x−22)(x−2)=0

⇒ x=225,2 x = \dfrac{22}{5}, 2x=522​,2

Substituting x=225x = \dfrac{22}{5}x=522​ in the given equation, we get

275+125=35\sqrt{\dfrac{27}{5}} + \sqrt{\dfrac{12}{5}} = \sqrt{\dfrac{3}{5}}527​​+512​​=53​​ ⇒ 27+12=3 \sqrt{27} + \sqrt{12} = \sqrt{3}27​+12​=3​

As all three square roots result in non-negative real numbers, the above equation is not possible. Therefore, x=225x = \dfrac{22}{5}x=522​ is not a solution.

Substituting x=2x = 2x=2 in the given equation, we get
3+0=3\sqrt{3} + \sqrt{0} = \sqrt{3}3​+0​=3​

This is true. Therefore, x=2x = 2x=2 is the only solution.

Answer: 222


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