Math, asked by Amayra1440, 1 month ago

If the values ​​p = 1 and q = –2 are the roots of the quadratic equation then the quadratic equation is *

1️⃣ x² + 2x –1 = 0
2️⃣ x² - x - 2 = 0
3️⃣ x² - 2x + 1 = 0
4️⃣ x² + x + 2 = 0​

Answers

Answered by PopularAnswerer01
92

Question:-

  • If the values p = 1 and q = –2 are the roots of the quadratic equation then the quadratic equation is

To Find:-

  • Find the equation.

Solution:-

Formula to be Used:-

  • \sf \: { x }^{ 2 } - ( \alpha + \beta )x + \alpha \beta = 0

Substitute the value of p in the equation:-

\sf\longrightarrow \: { x }^{ 2 } - ( p + q )x + pq = 0

\sf\longrightarrow \: { x }^{ 2 } - ( 1 + ( - 2 ) )x + ( 1 )( - 2 ) = 0

\sf\longrightarrow \: { x }^{ 2 } - ( 1 - 2 )x - 2 = 0

\sf\longrightarrow \: { x }^{ 2 } - ( - 1 )x - 2 = 0

\sf\longrightarrow \: { x }^{ 2 } + x - 2 = 0

Hence ,

  • \sf \: The \: equation \: is \: { x }^{ 2 } + x - 2 = 0
Answered by CuteAnswerer
27

GIVEN :

  • p = 1 and q= - 2 are the roots of the quadratic equation.

TO FIND :

  • Quadratic equation.

FORMULA REQUIRED :

  •  \underline{\boxed{\purple{\bf{x^2- (\alpha + \beta)x + \alpha\beta =0 }}}}

SOLUTION :

We have,

  • p = 1

  • q = - 2

Hence, we can write as :

  • \bf x^2- (p + q)x + pq = 0

By substituting the value of p = 1 and q = -2 we get,

\implies {\sf x^2 -  [1 + ( - 2)]x + 1 \times (- 2) = 0}\\ \\

\implies {\sf x^2- (1 - 2)x  - 2 =0 }\\  \\

\implies {\sf x^2- (- 1)x  - 2  =0 }\\  \\

\implies{ \underline{ \boxed{ \red{\bf{ x^2 + x - 2 =0}}}}}

\huge{\green{\therefore}} Quadratic equation = x² + x - 2 = 0.

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