Question

2x²-5x-7) (x-1) =0 ​

Answer 1

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Answer 2

To solve the given quadratic equation $(2x^2 - 5x - 7)(x - 1) = 0$, you can use the zero-product property, which states that if the product of several factors is equal to zero, then at least one of the factors must be zero. So, you can set each factor equal to zero and solve for $x$.

Setting $2x^2 - 5x - 7$ equal to zero:

$2x^2 - 5x - 7 = 0$

This quadratic equation can be factored, or you can use the quadratic formula to find the solutions. If you use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the quadratic equation $2x^2 - 5x - 7 = 0$, where $a = 2$, $b = -5$, and $c = -7$, the solutions would be:

$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(-7)}}{2(2)}$ $x = \frac{5 \pm \sqrt{89}}{4}$

So the two solutions for this quadratic equation are:

$x = \frac{5 + \sqrt{89}}{4}$ $x = \frac{5 - \sqrt{89}}{4}$

Setting $x - 1$ equal to zero:

$x - 1 = 0$

Solving for $x$, you get:

$x = 1$

Therefore, the solutions to the given equation $(2x^2 - 5x - 7)(x - 1) = 0$ are $x = 1$, $x = \frac{5 + \sqrt{89}}{4}$, and $x = \frac{5 - \sqrt{89}}{4}$.