Question

the roots of a quadratic equation are 5 and -2. find the equation

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Answer 1

$\huge{\bf{\green{\mathfrak{Question:-}}}}$

• The roots of a quadratic equation are 5 and -2. Find the equation.

$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

• $\textsf{The roots of the quadratic equation are 5 and - 2.}$

• $\boxed{\therefore{\sf{Factors \: of \: the \: quadratic \: equation \: will \: be \: (x \: - \: 5) \: and \: (x \: + \: 2).}}}$

• $\textsf{By multiplying the factors we get the Quadratic equation,}$

• $\sf{(x \: - \: 5) \: (x \: + \: 2)}$

• $\sf{x \: (x \: + \: 2) \: - \: 5(x \: + \: 2)}$

• $\sf{x^{2} \: + \: 2x \: - \: 5x \: - \: 10}$

• $\boxed{\sf{x^{2} \: - \: 3x \: - \: 10}}$

$\huge{\bf{\red{\mathfrak{Conclusion:-}}}}$

• $\boxed{\therefore{\sf{x^{2} \: - \: 3x \: - \: 10 \: is \: the \: quadratic \: equation \: whose \: roots \: are \: 5 \: and \: - \: 2.}}}$

$\huge{\bf{\pink{\mathfrak{Verification:-}}}}$

• $\sf{x^{2} \: - \: 3x \: - \: 10 \: = \: 0}$

• $\textsf{Substituting x = 5,}$

• $\sf{5^{2} \: - 3 \: * \: 5 \: - \: 10 \: = \: 0}$

• $\sf{25 \: - \: 15 \: - \: 10 \: = \: 0}$

• $\sf{10 \: - \: 10 \: = \: 0}$

• $\sf{0 \: = \: 0}$

• $\textsf{Substituting x = - 2,}$

• $\sf{- \: 2^{2} \: - 3 \: * \: - \: 2 \: - \: 10 \: = \: 0}$

• $\sf{4 \: + \: 6 \: - \: 10 \: = \: 0}$

• $\sf{10 \: - \: 10 \: = \: 0}$

• $\sf{0 \: = \: 0}$

• $\textsf{LHS = RHS in both the cases.}$

$\huge{\bf{\purple{\mathfrak{Note:-}}}}$

• If roots of a quadratic equation is given, then the factors of that quadratic equation can be found out by equating the roots to x.

• $\textsf{In the above problem,}$

• $\sf{5 \:= \:x}$

• x - 5 = 0 is the factor of the Quadratic equation.

• $\sf{- \:2 \:= \:x}$

• x + 2 = 0 is the factor of the Quadratic equation.

• Therefore, (x - 5) and (x + 2) are the factors of the Quadratic equation.

• Ignore ( = 0) as we require here only the factors.

Answer 2

Question:-

• The roots of a quadratic equation are 5 and -2. find the equation.

To Find:-

• Find the equation.

Given:-

• The roots of the equation are 5 and -2.

Solution:-

Formula Used:-

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

Substitute Values:-

• The value of α is 5.

• The value of β is -2.

$\tt\implies \: { x }^{ 2 } - [ 5 + ( -2 ) ]x + ( 5 )( -2 ) = 0$

$\tt\implies \: { x }^{ 2 } - ( 5 - 2 )x + ( -10 ) = 0$

$\tt\implies \: { x }^{ 2 } - 3x - 10 = 0$

Verification:-

Substitute x = 5:-

$\tt\implies \: { x }^{ 2 } - 3x - 10 = 0$

$\tt\implies \: { 5 }^{ 2 } - 3( 5 ) - 10 = 0$

$\tt\implies \: 25 - 15 - 10 = 0$

$\tt\implies \: 25 - 25 = 0$

$\tt\implies \: 0 = 0$