What are the solutions to the equation (x-a+2)(x-a+3)=42 ?

The solutions to the equation (x-a+2)(x-a+3)=42 are x=a+4,x=a-9

Find the zeros of (x-a+2)(x-a+3)=42

The zeros of (x-a+2)(x-a+3)=42 are x=a+4,x=a-9

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(x - a + 2) (x - a + 3) = 42

x = a + 4, x = a - 9

Solution steps

(x - a + 2) (x - a + 3) = 42

Expand

(x - a + 2) (x - a + 3) : - 2 a x + x^{2} + 5 x + a^{2} - 5 a + 6

- 2 a x + x^{2} + 5 x + a^{2} - 5 a + 6 = 42

Move

42

to the left side

x^{2} + 5 x - 2 a x + a^{2} - 5 a - 36 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + (5 - 2 a) x + a^{2} - 5 a - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 5 - 2 a ) \pm ( 5 - 2 a ) ^{2} - 4 \cdot 1 \cdot ( a ^{2} - 5 a - 36 )}{2 \cdot 1}

(5 - 2 a)^{2} - 4 \cdot 1 \cdot (a^{2} - 5 a - 36) = 13

x_{1, 2} = \frac{- ( 5 - 2 a ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 5 - 2 a ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( 5 - 2 a ) - 13}{2 \cdot 1}

x = \frac{- ( 5 - 2 a ) + 13}{2 \cdot 1} : a + 4

x = \frac{- ( 5 - 2 a ) - 13}{2 \cdot 1} : a - 9

The solutions to the quadratic equation are:

x = a + 4, x = a - 9

- 2 x^{2} + x + 2 = 0

x^{2} - 2 x + 8

20 x^{2} - 7 x = 6

(x - 4)^{2} - 20 = 0

x (x - 1) = 3

What are the solutions to the equation (x-a+2)(x-a+3)=42 ?
The solutions to the equation (x-a+2)(x-a+3)=42 are x=a+4,x=a-9
Find the zeros of (x-a+2)(x-a+3)=42
The zeros of (x-a+2)(x-a+3)=42 are x=a+4,x=a-9