Question

X^2+4x-(a^2+2a-3)=0
solve then quadratic equation by factorisation method.

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO SOLVE

Using factorisation method

$\sf{ {x}^{2} + 4x - ( {a}^{2} + 2a - 3) \: } = 0$

EVALUATION

$\sf{ {x}^{2} + 4x - ( {a}^{2} + 2a - 3) \: } = 0$

$\implies \sf{ {x}^{2} + 4x - ( {a}^{2} + 3a - a- 3) \: } = 0$

$\implies \sf{ {x}^{2} + 4x - \bigg [ {a}(a + 3) - 1(a + 3) \bigg] \: } = 0$

$\implies \sf{ {x}^{2} + 4x - (a + 3) (a - 1) \: } = 0$

$\implies \sf{ {x}^{2} + \bigg[ (a + 3) - (a - 1) \bigg] x - (a + 3) (a - 1) \: } = 0$

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

$\implies \sf{ {x}(x+ a + 3) - (a - 1) (x + a + 3) \: } = 0$

$\implies \sf{ (x+ a + 3) (x - a + 1)\: } = 0$

Since the product of two real numbers are zero then either of them are zero

$\implies \sf{ (x+ a + 3) = 0 \: \: \: or \: \: \: (x - a + 1)\: } = 0$

Now

$\sf{ (x+ a + 3) = 0 \:} \: \: \: gives$

$\sf{ x = - ( a + 3) \:} \: \: \:$

Again

$\sf{ (x - a + 1) = 0 \:} \: \: \: gives$

$\sf{ x = a - 1 \:} \: \: \:$

Hence the required solution is

$\sf{x = - (a+3 ) \: \: , \: \: (a-1)}$

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LEARN MORE FROM BRAINLY

Let A={1,2,0,-1} B={1,3-1,-3} and

function f = A to B f(x) =px+q

If f={ (1,1),(2,3),(0,-1),(-1,-3)} find the value of p

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