Question

Please solve question 17. Best answer will be marked as Brainliest. Answer is 2 and -7/2. Please solve it with method.​

Answer 1

see in the attachment.

Answer 2

Easier method :

1. Let $t=2x^2+3x$ for convenience.
2. Leave only one square root in LHS.
3. Repeat 2 until there is no square root.
4. Substitute $t=2x^2+3x$ and solve the equation.
5. Check the root.

Step 1

$\sqrt{t+2} +\sqrt{t+11} =9$

Step 2

$(\sqrt{t+2} )^2=(9-\sqrt{t+11} )^2$

$t+2=81-18\sqrt{t+11}+(t+11)$

$18\sqrt{t+11} =90$

$(\sqrt{t+11} )^2=5^2$

Step 3

$t+11=25$

Step 4

$2x^2+3x-14=0$

$(2x+7)(x-2)=0$

∴ $x=-\frac{7}{2}$ or $x=2$

Step 5

Use $t=2x^2+3x$ in convenience.

Whether $x=-\frac{7}{2}$ or $x=2$, it makes $t$ satisfy the expression $t-14=0$.

We know that $t=14$ now.

Substitute into the square root.

$\sqrt{14+2}+\sqrt{14+11} =9$

∴Two solutions $x=-\frac{7}{2}$, $x=2$ satisfies the equation.