Question

Determine the set of values of k for which the equation (3-2k)x^2 + (2k-3)x+1=0 has no real roots

Answer 1

EXPLANATION.

Set of value of k for which the equation,

⇒ (3 - 2k)x² + (2k - 3)x + 1 = 0.

As we know that,

No real roots = D < 0.

⇒ b² - 4ac < 0.

⇒ (2k - 3)² - 4[(3 - 2k)(1)] < 0.

⇒ 4k² + 9 - 12k - 4[3 - 2k] < 0.

⇒ 4k² + 9 - 12k - 12 + 8k < 0.

⇒ 4k² - 4k - 3 < 0.

Factorizes into middle term split, we get.

⇒ 4k² - 6k + 2k - 3 < 0.

⇒ 2k(2k - 3) + 1(2k - 3) < 0.

⇒ (2k + 1)(2k - 3) < 0.

Plot the line in a wavy curve method, we get.

As we know that,

Find the zeroes of the equation, we get.

⇒ (2k + 1) = 0.

⇒ k = -1/2.

⇒ (2k - 3) = 0.

⇒ k = 3/2.

Plot in wavy curve method, we get.

⇒ -1/2 < k < 3/2.

⇒ k ∈ (-1/2, 3/2).

                                                                                                                     

MORE INFORMATION.

Location of roots of a quadratic equation ax² + bx + c = 0.

(1) = Conditions for both the roots will be greater than k.

(a) = D ≥ 0.

(b) = k < -b/2a.

(c) = af(k) > 0.

(2) = Conditions for both the roots will be less than k.

(a) = D ≥ 0.

(b) = k > -b/2a.

(c) = af(k) > 0.

(3) = Conditions for k lie between the roots.

(a) = D > 0.

(b) = af(k) < 0.

(4) = Conditions for exactly one root lie in the interval (k₁, k₂) where k₁ < k₂.

(a) = f(k₁). f(k₂) < 0.

(b) = D > 0.

(5) = When both roots lie in the interval (k₁, k₂) where k₁ < k₂.

(a) = D > 0.

(b) = f(k₁). f(k₂) > 0.

(6) = Any algebraic expression f(x) = 0 in interval [a, b] if.

(a) = Sign of f(a) and f(b) are of same then either no roots or even no. of roots exist.

(b) = Sign of f(a) and f(b) are opposite then f(x) = 0 has at least one real root or odd no. of roots.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• Determine the set of values of k for which the equation has no real roots.

$\sf \: (3 - 2k) {x}^{2} + (2k - 3)x + 1 = 0$

$\large\underline\purple{\bold{Solution :- }}$

Given

Quadratic equation is

$\rm : \implies \: \: (3 - 2k) {x}^{2} + (2k - 3)x + 1 = 0$

$\rm : \implies \:On \: comparing \: with \: {ax}^{2} +bx + c= 0, \: we \: get$

$\rm : \implies \:a \: = \: 3 - 2k$

$\rm : \implies \:b \: = \: 2k - 3$

$\rm : \implies \:c \: = \: 1$

Since,

• Given Quadratic equation has no real roots

So,

$\boxed{ \pink{ \rm \: Discriminant, \: D \: < \: 0}}$

$\rm : \implies \: {b}^{2} - 4ac < 0$

$\rm : \implies \: {(2k - 3)}^{2} - 4 \times 1 \times (3 - 2k) \: < \: 0$

$\rm : \implies \: {(2k - 3)}^{2} + 4(2k - 3) < 0$

$\rm : \implies \:(2k - 3)(2k - 3 + 4) < 0$

$\rm : \implies \:(2k - 3)(2k + 1) < 0$

$\rm : \implies \:\boxed{ \pink{ \rm\: - \dfrac{1}{2} < k < \dfrac{3}{2} }}$

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Explore more :-

A quadratic inequality is an equation of second degree that uses an inequality sign instead of an equal sign.

The solutions to quadratic inequality always give the two roots. The nature of the roots may differ and can be determined by discriminant (b2 – 4ac).

The general forms of the quadratic inequalities are:

• ax^2 + bx + c < 0
• ax^2 + bx + c ≤ 0
• ax^2 + bx + c > 0
• ax^2 + bx + c ≥ 0

To solve a quadratic inequality, we also apply the same method as illustrated in the procedure below:

• Write the quadratic inequality in standard form: ax2 + bx + c where a, b and are coefficients and a ≠ 0
• Determine the roots of the inequality.
• Write the solution in inequality notation or interval notation.
• If the quadratic inequality is in the form: (x – a) (x – b) ≤ 0, then a ≤ x ≤ b,
• if it is in the form :(x – a) (x – b) ≥ 0, when a < b then a ≤ x or x ≥ b.
• If the quadratic inequality is in the form: (x – a) (x – b) < 0, then a < x < b,
• if it is in the form :(x – a) (x – b) > 0, when a < b then a <x or x > b.