Question

Which of the following values of \[y\] satisfies the equation?
\[2{\left( {{y^2} - 6y} \right)^2} - 8\left( {{y^2} - 6y + 3} \right) - 40 = 0\]
A. \[y = - 3 \pm \sqrt {17} \]
B. \[y = 8 \pm \sqrt 5 \]
C. \[y = - 8 \pm \sqrt 5 \]
D. \[y = 3 \pm \sqrt {17} \]

Answer 1

Hint: We have to find the value of \[y\] satisfies the equation \[2{\left( {{y^2} - 6y} \right)^2} - 8\left( {{y^2} - 6y + 3} \right) - 40 = 0\]. For this, we will assume \[{y^2} - 6y = t\]. Then we will solve the quadratic equation in \[t\]. After this, we will substitute back \[{y^2} - 6y = t\] and we will solve the quadratic equation in \[y\] to find the value of \[y\] that satisfies the given equation.

Complete step by step answer:
We have to find the value of \[y\] satisfies the equation \[2{\left( {{y^2} - 6y} \right)^2} - 8\left( {{y^2} - 6y + 3} \right) - 40 = 0\].
Let \[{y^2} - 6y = t\].
Putting this in the given equation, we get
\[ \Rightarrow 2{\left( t \right)^2} - 8\left( {t + 3} \right) - 40 = 0\]
On simplifying, we get
\[ \Rightarrow 2{t^2} - 8t - 24 - 40 = 0\]
\[ \Rightarrow 2{t^2} - 8 - 64 = 0\]

On dividing both the sides \[2\], we get
\[ \Rightarrow {t^2} - 4t - 32 = 0\]
As we know, for a quadratic equation, \[a{x^2} + bx + c\], where \[a\], \[b\] and \[c\], roots is given by: \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Using this, we get the roots of \[{t^2} - 4t - 32 = 0\] as,
\[ \Rightarrow t = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times \left( 1 \right) \times \left( { - 32} \right)} }}{{2 \times 1}}\]
\[ \Rightarrow t = \dfrac{{4 \pm \sqrt {16 + 128} }}{2}\]

On simplifying, we get
\[ \Rightarrow t = \dfrac{{4 \pm \sqrt {144} }}{2}\]
\[ \Rightarrow t = \dfrac{{4 \pm 12}}{2}\]
So, we can write,
\[ \Rightarrow t = \dfrac{{4 + 12}}{2}\] or \[t = \dfrac{{4 - 12}}{2}\]
\[ \Rightarrow t = \dfrac{{16}}{2}\] or \[t = \dfrac{{ - 8}}{2}\]
On simplifying, we get
\[ \Rightarrow t = 8\] or \[t = - 4\]
When \[t = 8\], substituting back \[{y^2} - 6y = t\], we get
\[ \Rightarrow {y^2} - 6y = 8\]
\[ \Rightarrow {y^2} - 6y - 8 = 0\]

On solving, we get
\[ \Rightarrow y = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4 \times 1 \times \left( { - 8} \right)} }}{{2 \times 1}}\]
\[ \Rightarrow y = \dfrac{{6 \pm \sqrt {36 + 32} }}{2}\]
On simplifying, we get
\[ \Rightarrow y = \dfrac{{6 \pm \sqrt {68} }}{2}\]
\[ \Rightarrow y = \dfrac{{6 \pm 2\sqrt {17} }}{2}\]
On further simplification, we get
\[ \Rightarrow y = 3 \pm \sqrt {17} \]
Now, when \[t = - 4\], substituting back \[{y^2} - 6y = t\], we get
\[ \Rightarrow {y^2} - 6y = - 4\]
\[ \Rightarrow {y^2} - 6y + 4 = 0\]

On solving, we get
\[ \Rightarrow y = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4 \times 1 \times 4} }}{{2 \times 1}}\]
\[ \Rightarrow y = \dfrac{{6 \pm \sqrt {36 - 16} }}{2}\]
On simplifying, we get
\[ \Rightarrow y = \dfrac{{6 \pm \sqrt {20} }}{2}\]
\[ \Rightarrow y = \dfrac{{6 \pm 2\sqrt 5 }}{2}\]
On further simplification, we get
\[ \Rightarrow y = 3 \pm \sqrt 5 \]
Therefore, from the given options, the value of \[y\] which satisfies the equation \[2{\left( {{y^2} - 6y} \right)^2} - 8\left( {{y^2} - 6y + 3} \right) - 40 = 0\] is \[y = 3 \pm \sqrt {17} \].

Hence, option D is correct.

Note: In this question, we have used the concept of quadratic equations. A quadratic equation function may have one, two, or zero roots. Roots are also called the x-intercept or zeroes. Also, the y-coordinate of any points lying on the x-axis is zero. So, to find the roots of a quadratic function, we set \[f(x) = 0\].