Select the two values of x that are roots of this equation. 2x+11x+15= 0

Mathematics

1 answer:

Amiraneli [1.4K]3 years ago

5 0

Answer:

The roots of the equation are x=-3 and x=-2.5

Step-by-step explanation:

<u><em>The correct quadratic equation is</em></u>

2x^2+11x+15=0

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$2x^{2} +11x+15=0$

$a=2\\b=11\\c=15$

substitute in the formula

$x=\frac{-11(+/-)\sqrt{11^{2}-4(2)(15)}} {2(2)}$

$x=\frac{-11(+/-)\sqrt{121-120}} {4}$

$x=\frac{-11(+/-)\sqrt{1}} {4}$

$x=\frac{-11(+/-)1} {4}$

$x_1=\frac{-11(+)1}{4}=-2.5$

$x_2=\frac{-11(-)1}{4}=-3$

therefore

The roots of the equation are x=-3 and x=-2.5

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Is X=[−14−32] the inverse of A=[15−25310−110]?

Elis [28]

The product of the matrices is an identity matrix. Therefore, X and A are inverse of each other.

The matrices are given as:

$X = \left[\begin{array}{cc}-1&-3\\4&2\end{array}\right]$

$A = \left[\begin{array}{cc}\frac 15&\frac 3{10}\\-\frac 25 &-\frac 1{10}\end{array}\right]$

To check the matrices are inverse, we calculate their products.

$A \times X = \left[\begin{array}{cc}\frac 15&\frac 3{10}\\-\frac 25 &-\frac 1{10}\end{array}\right] \times \left[\begin{array}{cc}-1&-3\\4&2\end{array}\right]$

Multiply the rows of A by the column of X.

This gives

$A \times X = \left[\begin{array}{cc}\frac 15 \times -1 + \frac{3}{10} \times 4&\frac 15 \times -3 + \frac{3}{10} \times 2\\ -\frac 25 \times -1 - \frac{1}{10} \times 4&-\frac 25 \times -3 + -\frac{1}{10} \times 2\end{array}\right]$