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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 - Changes to Roots

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3.5 Changes to Roots

Where α\alphaα and β\betaβ are the roots of a quadratic equation ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0, and kkk is a constant,

For the roots The quadratic equation is
(α+k)(\alpha + k)(α+k) and (β+k)(\beta + k)(β+k) a(x−k)2+b(x−k)+c=0a(x - k)^{2} + b(x - k) + c = 0a(x−k)2+b(x−k)+c=0
(α−k)(\alpha - k)(α−k) and (β−k)(\beta - k)(β−k) a(x+k)2+b(x+k)+c=0a(x + k)^{2} + b(x + k) + c = 0a(x+k)2+b(x+k)+c=0
(αk)(\alpha k)(αk) and (βk)(\beta k)(βk) a(xk)2+b(xk)+c=0a \left(\dfrac{x}{k} \right)^{2} + b \left(\dfrac{x}{k} \right) + c = 0a(kx​)2+b(kx​)+c=0
(αk)\left(\dfrac{\alpha}{k} \right)(kα​) and (βk)\left(\dfrac{\beta}{k} \right)(kβ​) a(xk)2+b(xk)+c=0a(xk)^{2} + b(xk) + c = 0a(xk)2+b(xk)+c=0
(1α)\left(\dfrac{1}{\alpha} \right)(α1​) and (1β)\left(\dfrac{1}{\beta} \right)(β1​) cx2+bx+a=0cx^{2} + bx + a = 0cx2+bx+a=0


Ratio of Roots: For a quadratic equation ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0, if the ratio of roots, i.e., α:β=p:q\alpha : \beta = p : qα:β=p:q, then b2pq=ac(p+q)2\bm{b^{2}pq = ac(p + q)^{2}}b2pq=ac(p+q)2

Example 9

If α\alphaα and β\betaβ are the roots of the quadratic equation 2x2+3x−4=02x^{2} + 3x - 4 = 02x2+3x−4=0, what is the equation whose roots are
(I) α+52\dfrac{\alpha + 5}{2}2α+5​ and β+52\dfrac{\beta + 5}{2}2β+5​ ?
(II) two-third the reciprocal of α\alphaα and β\betaβ?

Solution

Case I: Equation whose roots are α\alphaα and β\betaβ ⇒ 2x2+3x−4=0 2x^{2} + 3x - 4 = 02x2+3x−4=0

Equation whose roots are (α+5)(\alpha + 5)(α+5) and (β+5)(\beta + 5)(β+5) ⇒ 2(x−5)2+3(x−5)−4=0 2(x - 5)^{2} + 3(x - 5) - 4 = 02(x−5)2+3(x−5)−4=0
⇒ 2x2−17x+31=0 2x^{2} - 17x + 31 = 02x2−17x+31=0

Equation whose roots are (α+52)\left(\dfrac{\alpha + 5}{2} \right)(2α+5​) and (β+52)\left(\dfrac{\beta + 5}{2} \right)(2β+5​) ⇒ 2(2x)2−17(2x)+31=0 2(2x)^{2} - 17(2x) + 31 = 02(2x)2−17(2x)+31=0

⇒ 8x2−34x+31=0 8x^{2} - 34x + 31 = 08x2−34x+31=0

Case II: Equation whose roots are 1α\dfrac{1}{\alpha}α1​ and 1β\dfrac{1}{\beta}β1​ ⇒ −4x2+3x+2=0 -4x^{2} + 3x + 2 = 0−4x2+3x+2=0

⇒ 4x2−3x−2=0 4x^{2} - 3x - 2 = 04x2−3x−2=0

Equation whose roots are (23α)\left(\dfrac{2}{3\alpha} \right)(3α2​) and (23β)\left(\dfrac{2}{3\beta} \right)(3β2​) ⇒ 4(32×x)−2=0 4 \left(\dfrac{3}{2} \times x \right) - 2 = 04(23​×x)−2=0

⇒ 18x2−9x−4=0 18x^{2} - 9x - 4 = 018x2−9x−4=0

Answer: (I) 8x2−34x+31=08x^{2} - 34x + 31 = 08x2−34x+31=0; (II) 18x2−9x−4=018x^{2} - 9x - 4 = 018x2−9x−4=0


Example 10

In the equation 4x2−12x+m=04x^{2} - 12x + m = 04x2−12x+m=0, if one root is 555 times the other, then m=m =m= ?

Solution

Ratio of roots =p:q=1:5= p : q = 1 : 5=p:q=1:5

b2pq=ac(p+q)2b^{2}pq = ac(p + q)^{2}b2pq=ac(p+q)2
⇒ (−12)2×1×5=4m(1+5)2 (-12)^{2} \times 1 \times 5 = 4m(1 + 5)^{2}(−12)2×1×5=4m(1+5)2
⇒ 144×5=4m×36 144 \times 5 = 4m \times 36144×5=4m×36
⇒ m=5 m = 5m=5

Answer: 555


3.6 Applying Binomial Identities to find new Equation

Example 11

If α\alphaα and β\betaβ are the roots of the equation x2+5x−3=0x^{2} + 5x - 3 = 0x2+5x−3=0, what is the equation whose roots are
(I) 1α2\dfrac{1}{\alpha^{2}}α21​ and 1β2\dfrac{1}{\beta^{2}}β21​?

(II) α2β\dfrac{\alpha^{2}}{\beta}βα2​ and β2α\dfrac{\beta^{2}}{\alpha}αβ2​?

Solution

Sum of roots =α+β=−5= \alpha + \beta = -5=α+β=−5
Product of roots =αβ=−3= \alpha \beta = -3=αβ=−3

Case I: Where roots are 1α2\dfrac{1}{\alpha^{2}}α21​ and 1β2\dfrac{1}{\beta^{2}}β21​

Sum of roots =1α2+1β2=α2+β2(α2β2)= \dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} = \dfrac{\alpha^{2} + \beta^{2}}{(\alpha^{2} \beta^{2})}=α21​+β21​=(α2β2)α2+β2​ =(α+β)2−2αβ(αβ)2=(−5)2−2×−3(−3)2= \dfrac{(\alpha + \beta)^{2} - 2 \alpha \beta}{(\alpha \beta)^{2}} = \dfrac{(-5)^{2} - 2 \times -3}{(-3)^{2}}=(αβ)2(α+β)2−2αβ​=(−3)2(−5)2−2×−3​

=319= \dfrac{31}{9}=931​

Product of roots =1α2×1β2=1(αβ)2=19= \dfrac{1}{\alpha^{2}} \times \dfrac{1}{\beta^{2}} = \dfrac{1}{(\alpha \beta)^{2}} = \dfrac{1}{9}=α21​×β21​=(αβ)21​=91​

New Equation ⇒ x2−319x+19=0 x^{2} - \dfrac{31}{9}x + \dfrac{1}{9} = 0x2−931​x+91​=0 ⇒ 9x2−31x+1=0 9x^{2} - 31x + 1 = 09x2−31x+1=0

Case II: Where roots are α2β\dfrac{\alpha^{2}}{\beta}βα2​ and β2α\dfrac{\beta^{2}}{\alpha}αβ2​

Sum of roots =α2β+β2α=α3+β3αβ= \dfrac{\alpha^{2}}{\beta} + \dfrac{\beta^{2}}{\alpha} = \dfrac{\alpha^{3} + \beta^{3}}{\alpha \beta}=βα2​+αβ2​=αβα3+β3​ =(α+β)3−3αβ(α+β)αβ= \dfrac{(\alpha + \beta)^{3} - 3 \alpha \beta (\alpha + \beta)}{\alpha \beta}=αβ(α+β)3−3αβ(α+β)​ =(−5)3−3×−3×−5−3=1703= \dfrac{(-5)^{3} - 3 \times -3 \times -5}{-3} = \dfrac{170}{3}=−3(−5)3−3×−3×−5​=3170​

Product of roots =α2β×β2α=αβ=−3= \dfrac{\alpha^{2}}{\beta} \times \dfrac{\beta^{2}}{\alpha} = \alpha \beta = -3=βα2​×αβ2​=αβ=−3

New Equation ⇒ x2−1703x−3=0 x^{2} - \dfrac{170}{3}x - 3 = 0x2−3170​x−3=0 ⇒ 3x2−170x−9=0 3x^{2} - 170x - 9 = 03x2−170x−9=0

Answer: (I) 9x2−31x+1=09x^{2} - 31x + 1 = 09x2−31x+1=0; (II) 3x2−170x−9=03x^{2} - 170x - 9 = 03x2−170x−9=0


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