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If |2 3 -1| = -106, find the value of x.

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55. If ∣23−15x3−352∣=−106\begin{vmatrix} 2 & 3 & -1 \\ 5 & x & 3 \\ -3 & 5 & 2 \end{vmatrix} = -106 find the value of xx.

  1. Expand the determinant.

    2(x⋅2−3⋅5)−3(5⋅2−3⋅(−3))+(−1)(5⋅5−x⋅(−3))=−1062(x \cdot 2 - 3 \cdot 5) - 3(5 \cdot 2 - 3 \cdot (-3)) + (-1)(5 \cdot 5 - x \cdot (-3)) = -106

  2. Simplify the expression.

    2(2x−15)−3(10+9)−1(25+3x)=−1062(2x - 15) - 3(10 + 9) - 1(25 + 3x) = -106 4x−30−3(19)−25−3x=−1064x - 30 - 3(19) - 25 - 3x = -106 4x−30−57−25−3x=−1064x - 30 - 57 - 25 - 3x = -106

  3. Combine like terms.

    (4x−3x)+(−30−57−25)=−106(4x - 3x) + (-30 - 57 - 25) = -106 x−112=−106x - 112 = -106

  4. Solve for xx.

    x=−106+112x = -106 + 112 x=6x = 6

    The correct option is A.
    The value of xx is 6\boxed{\text{6}}.

    56. Solve the equation 2x2−32=02x^2 - 32 = 0.

  5. Isolate the x2x^2 term.

    2x2=322x^2 = 32

  6. Divide by 2.

    x2=322x^2 = \frac{32}{2} x2=16x^2 = 16

  7. Take the square root of both sides.

    x=±16x = \pm\sqrt{16} x=±4x = \pm 4

    The correct option is B.
    The solutions are x=4 or x=−4\boxed{x = 4 \text{ or } x = -4}.

    57. Find the sum and the product of the roots of the equation 3x2−2x−6=03x^2 - 2x - 6 = 0 respectively.

  8. Identify the coefficients a,b,ca, b, c from the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.

    For 3x2−2x−6=03x^2 - 2x - 6 = 0, we have a=3a = 3, b=−2b = -2, c=−6c = -6.

  9. Calculate the sum of the roots using the formula −ba-\frac{b}{a}.

    Sum of roots=−−23=23\text{Sum of roots} = -\frac{-2}{3} = \frac{2}{3}

  10. Calculate the product of the roots using the formula ca\frac{c}{a}.

    Product of roots=−63=−2\text{Product of roots} = \frac{-6}{3} = -2

    The correct option is B.
    The sum of the roots is 23\boxed{\frac{2}{3}} and the product of the roots is −2\boxed{-2}.

    58. Given that P=(−2134)P = \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix} and Q=(5−32−1)Q = \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix}, find the value of 2P+Q2P + Q.

  11. Calculate 2P2P by multiplying each element of matrix PP by 2.

    2P=2(−2134)=(2×(−2)2×12×32×4)=(−4268)2P = 2 \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 \times (-2) & 2 \times 1 \\ 2 \times 3 & 2 \times 4 \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 6 & 8 \end{pmatrix}

  12. Add matrix 2P2P and matrix QQ by adding their corresponding elements.

    2P+Q=(−4268)+(5−32−1)2P + Q = \begin{pmatrix} -4 & 2 \\ 6 & 8 \end{pmatrix} + \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix} 2P+Q=(−4+52+(−3)6+28+(−1))2P + Q = \begin{pmatrix} -4 + 5 & 2 + (-3) \\ 6 + 2 & 8 + (-1) \end{pmatrix}

  13. Perform the addition.

    2P+Q=(1−187)2P + Q = \begin{pmatrix} 1 & -1 \\ 8 & 7 \end{pmatrix}

    The correct option is E.
    The value of 2P+Q2P + Q is (1−187)\boxed{\begin{pmatrix} 1 & -1 \\ 8 & 7 \end{pmatrix}}.

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