Question

[Maths]

Find the value(s) of p for which the quadratic equations:
$\dag \: \: \bold\blue{ \tt(2p + 1) {x}^{2} - (7p + 3)x + (7p - 3) = 0}$
has equal roots. Also find these roots.​

Answer 1

Step-by-step explanation:

A quadratic equation has equal roots if it can be expressed as a "perfect square".

f(x) = x² + p(4x + p - 1) + 2

f(x) = x² + 4px + p² - p + 2

f(x) = x² + 4px + 4p² - (3p² + p - 2)

f(x) = (x + 2p)² - (3p² + p - 2)

For the above expression to be a perfect square,

3p² + p - 2 = 0

3p² + 3p - 2p - 2 = 0

3p(p + 1) - 2(p + 1) = 0

(p + 1)(3p - 2) = 0

=> p = -1 and ⅔

Answer 2

★ Concept :-

Here the concept of Discriminant and Splitting the Middle Term has been used. We see that we are given the equation where the roots of the equation are equal. So this means the discriminant of the equation will result as zero. Since we have two variables p and x in the equation so using discriminant we can find the value of p. Then after that we shall use this value of p to find the value of x.

Let's do it !!

______________________________________

★ Solution :-

Given,

» (2p + 1)x² - (7p + 2)x + (7p - 3) = 0

(2p + 1)x² - (7p + 2)x + (7p - 3) = 0This is the correct equation.

We are given that the roots of this equation are equal.

We know that, when

• D < 0 (roots are imaginary)

• D = 0 (roots are virtual)

• D > 0 (roots are real and distinct)

Here second condition follows.

From the given equation,

• a = (2p + 1)
• a = (2p + 1)b = - (7p + 2)
• a = (2p + 1)b = - (7p + 2)c = (7p - 3)

We know that,

→ D = b² - 4ac

Applying this in the second condition, we get

>> b² - 4ac = D

Here let's applybthe value of the terms.

>> [-(7p + 2)]² - 4(2p + 1)(7p - 3) = 0

Let's start solving this.

>> [(7p + 2)(7p + 2)] - 4[14p² - 6p + 7p - 3] = 0

Here we took square of -ve sign in first term as (- × - = +)

>> [49p² + 14p + 14p + 4] - 4[14p² + p - 3] = 0

>> 49p² + 28p + 4 - 56p² - 4p + 12 = 0

>> -7p² + 24p + 16 = 0

Taking all the terms to R.H.S., we get

>> 7p² - 24p - 16 = 0

• Now by using the method of Splitting the Middle Term, we get

>> 7p² - 28p + 4p - 16 = 0

On taking taking the common terms, we get

>> 7p(p - 4) + 4(p - 4) = 0

>> (7p + 4)(p - 4) = 0

Here either (7p + 4) = 0 or (p - 4) = 0.

So,

>> (7p + 4) = 0 or p - 4 = 0

>> p = -4/7 or p = 4

$\;\;\bf{\mapsto\;\;\red{p\;=\;\dfrac{-4}{7},\;4}}$

Now let's find the value of x using the value of p.

• Value of x when p = -4/7 ::

Let's apply the value of p in the initial equation.

$\;\tt{\rightarrow\;\:\bigg[2\bigg(\dfrac{-4}{7}\bigg)\:+\:1\bigg]x^{2}\:-\:\bigg[7\bigg(\dfrac{-4}{7}\bigg)\:+\:2\bigg]x\:+\:\bigg[7\bigg(\dfrac{-4}{7}\bigg)\:-\:3\bigg]\;=\;0}$

$\;\tt{\rightarrow\;\:\bigg[\dfrac{-8}{7}\:+\:1\bigg]x^{2}\:-\:\bigg[-4\:+\:2\bigg]x\:+\:\bigg[-4\:-\:3\bigg]\;=\;0}$

$\;\tt{\rightarrow\;\:\bigg[\dfrac{-8\;+\;7}{7}\bigg]x^{2}\:-\:\bigg[-4\:+\:2\bigg]x\:+\:\bigg[-4\:-\:3\bigg]\;=\;0}$

$\;\tt{\rightarrow\;\:\bigg[\dfrac{-1}{7}\bigg]x^{2}\:-\:\bigg[-2\bigg]x\:+\:\bigg[-7\bigg]\;=\;0}$

Taking -1/7 in common, we get

$\;\tt{\rightarrow\;\:\dfrac{-1}{7}\bigg[x^{2}\:-\:14x\:+\:49\bigg]\;=\;0}$

Then sending -1/7 to another side, we get

>> x² - 14x + 49 = 0

>> x² - 7x - 7x + 49 = 0

>> x(x - 7) - 7(x - 7) = 0

>> (x - 7)(x - 7) = 0

>> (x - 7)² = 0

>> x - 7 = 0 (since √0 = 0)

>> x = 7

Since the roots are equal. So,

• Hence x = 7, 7

• Value of x when p = 4 ::

Let's apply the value of p in the main equation. Then,

>> [2(4) + 1]x² - [7(4) + 2]x + [7(4) - 3] = 0

>> [8 + 1]x² - [28 + 2]x + [28 - 3] = 0

>> 9x² - 30x + 25 = 0

>> 9x² - 15x - 15x + 25 = 0

>> 3x(3x - 5) - 5(3x - 5) = 0

>> (3x - 5)(3x - 5) = 0

>> (3x - 5)² = 0

>> 3x - 5 = 0 (since √0 = 0)

>> 3x = 5

>> x = 5/3

Since the roots are equal. So,

• x = 5/3, 5/3

$\;\underline{\boxed{\tt{\purple{When\;p\;=\;\dfrac{-4}{7},\;then\;x\;=\;\bf{\green{7,\;7}}}}}}$

$\;\underline{\boxed{\tt{\purple{When\;p\;=\;4,\;then\;x\;=\;\bf{\pink{\dfrac{5}{3},\;\dfrac{5}{3}}}}}}}$