Question

find the value of m in which the quadratic equation has real or equal roots: (m+5)x²-(2m+3)x+(m-1)=0

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Answer 1

EXPLANATION.

Quadratic equation,

(m + 5)x² - (2m + 3)x + (m - 1) = 0.

As we know that,

D = Discriminant.  Or  b² - 4ac.

Equation has real and equal roots,

⇒ D = 0.

⇒ (2m + 3)² - 4(m + 5)(m - 1) = 0.

⇒ 4m² + 9 + 12m - 4[m² - m + 5m - 5] = 0.

⇒ 4m² + 9 + 12m - 4[m² + 4m - 5] = 0.

⇒ 4m² + 9 + 12m - 4m² - 16m + 20 = 0.

⇒ 9 + 12m - 16m + 20 = 0.

⇒ 29 - 4m = 0.

⇒ 4m = 29.

⇒  m = 29/4.

                                                                                                                         

MORE INFORMATION.

Quadratic expression.

A polynomial of degree two of the form ax² + bx + c (a ≠ 0) is called a quadratic expression in x.

The quadratic equation.

ax² + bx + c = 0 (a ≠ 0) has two roots, given by.

⇒ α = -b + √D/2a.

⇒ β = -b - √D/2a.

⇒ D = b² - 4ac.

Answer 2

Given Question :-

• find the value of m in which the quadratic equation has real or equal roots: (m+5)x²-(2m+3)x+(m-1)=0

Answer

$\begin{gathered}\begin{gathered}\bf \: Given \: -   \begin{cases} &\sf{Quadratic \: equation \: has \: real \: equal \: roots} \\ &\sf{(m + 5) {x}^{2} - (2m + 3)x + (m - 1) = 0 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  \: To \: Find  - \begin{cases} &\sf{the \: value \: of \: m}  \end{cases}\end{gathered}\end{gathered}$

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

The given Quadratic equation is

$\rm : \implies \:(m + 5) {x}^{2} - (2m + 3)x + (m - 1) = 0$

Now,

$\rm \:On \: comparing \: with \: a {x}^{2} +bx + c = 0 \: we \: get \:$

$\rm : \implies \:a \: = \: m \: + \: 5$

$\rm : \implies \:b \: = \: - \: (2m \: + \: 3)$

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

Since, it is given that equation has real and equal roots.

$\rm : \implies \:Discriminant \: (D) = \: 0$

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

$\rm : \implies \: { \bigg( - (2m + 3) \bigg)}^{2} - 4(m + 5)(m - 1) = 0$

$\rm : \implies \: {4m}^{2} + 9 + 12m -4 ( {m}^{2} - m + 5m - 5) = 0$

$\rm : \implies \:4 {m}^{2} + 9 + 12m - {4m }^{2} - 16m + 20 = 0$

$\rm : \implies \: - 4m + 29 = 0$

$\rm : \implies \:m \: = \: \dfrac{29}{4}$

Explore more :

Nature of roots of Quadratic Equation :-

The nature depends upon Discriminant, D. Three cases arises :

• D>0: When D is positive, the equation will have two real and distinct roots. This means the graph of the equation will intersect x-axis at exactly two different points.

• D = 0: When D is equal to zero, the equation will have two real and equal roots. This means the graph of the equation will intersect x-axis at exactly one point. 

• D < 0: When D is negative, the equation will have no real roots. This means the graph of the equation will not intersect x-axis.